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  2. 02Vectors
  3. 03Differentiation
  4. 04Applications of Differentiation
  5. 05Integration
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  8. 08Dimensional Formula
  9. 09Dimensional Analysis and Its Applications
  10. 10Experimental Skills

Kinematics

  1. 01Motion in One Dimension
  2. 02Motion in Multiple Dimensions
  3. 03Relative Velocity
  4. 04Circular Motion Kinematics
  5. 05Circular Motion Dynamics

Dynamics

  1. 01Forces and Laws of Motion
  2. 02Laws of Motion
  3. 03Friction
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  5. 05Fundamentals of Force
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  7. 07Applications: Objects in Equilibrium
  8. 08Applications: Objects in Motion
  9. 09Constraint Relations
  10. 10Inertial and Non-Inertial Frames
  11. 11Basics of Friction
  12. 12Applications of Friction

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  1. 01Conservation of Mechanical Energy
  2. 02Work and the Work-Energy Theorem
  3. 03Work and Kinetic Energy Theorem
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Theory/Kinematics

Kinematics · Chapter 02

Motion in Multiple Dimensions

Study of motion in two and three dimensions, including motion vectors, projectile motion, trajectory analysis, and range equations.

15 min read · 4 topics

01

Motion in Two and Three Dimensions

Introduction to Multi-Dimensional Motion

While motion in a straight line (one dimension) provides a foundation for understanding kinematics, most real-world motion occurs in two or three dimensions. To describe such motion, we must use vectors. These are mathematical objects that have both magnitude and direction.

A position vector r⃗\vec{r}r is used to specify the location of an object in a coordinate system. It is a vector drawn from the origin to the object's position.

In Cartesian coordinates:

  • For 2D motion (in a plane): r⃗=xi^+yj^\vec{r} = x\hat{i} + y\hat{j}r=xi^+yj^​
  • For 3D motion (in space): r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}r=xi^+yj^​+zk^

Here, i^\hat{i}i^, j^\hat{j}j^​, k^\hat{k}k^ are the unit vectors along the x, y, and z axes, respectively. The components (x, y, z) are the coordinates of the object.

A position vector in the 2D xy-plane
A position vector in the 2D xy-plane

A position vector in 3D space

True or False: A position vector in two dimensions requires three components to fully describe the location.

Displacement, Velocity, and Acceleration Vectors

The kinematic quantities we studied in 1D are now represented as vectors.

Displacement Vector

The displacement vector Δr⃗\Delta\vec{r}Δr represents the change in position from an initial position r1⃗\vec{r_1}r1​​ to a final position r2⃗\vec{r_2}r2​​.

Δr⃗=r2⃗−r1⃗\qquad \Delta\vec{r} = \vec{r_2} - \vec{r_1}Δr=r2​​−r1​​

In component form:

Δr⃗=(x2−x1)i^+(y2−y1)j^+(z2−z1)k^\qquad \Delta\vec{r} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}Δr=(x2​−x1​)i^+(y2​−y1​)j^​+(z2​−z1​)k^

The displacement vector is the difference between the final and initial position vectors.
The displacement vector is the difference between the final and initial position vectors.

Velocity Vector

The average velocity vector is the displacement vector divided by the time interval:

v⃗avg=Δr⃗Δt\qquad \vec{v}_{avg} = \dfrac{\Delta\vec{r}}{\Delta t}vavg​=ΔtΔr​

The instantaneous velocity vector is the derivative of the position vector with respect to time:

v⃗=dr⃗dt=dxdti^+dydtj^+dzdtk^=vxi^+vyj^+vzk^\qquad \vec{v} = \dfrac{d\vec{r}}{dt} = \dfrac{dx}{dt}\hat{i} + \dfrac{dy}{dt}\hat{j} + \dfrac{dz}{dt}\hat{k} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}v=dtdr​=dtdx​i^+dtdy​j^​+dtdz​k^=vx​i^+vy​j^​+vz​k^

The instantaneous velocity vector is always tangent to the path of the object.

Acceleration Vector

The average acceleration vector is the change in the velocity vector divided by the time interval:

a⃗avg=Δv⃗Δt\qquad \vec{a}_{avg} = \dfrac{\Delta\vec{v}}{\Delta t}aavg​=ΔtΔv​

The instantaneous acceleration vector is the derivative of the velocity vector with respect to time:

a⃗=dv⃗dt=dvxdti^+dvydtj^+dvzdtk^=axi^+ayj^+azk^\qquad \vec{a} = \dfrac{d\vec{v}}{dt} = \dfrac{dv_x}{dt}\hat{i} + \dfrac{dv_y}{dt}\hat{j} + \dfrac{dv_z}{dt}\hat{k} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}a=dtdv​=dtdvx​​i^+dtdvy​​j^​+dtdvz​​k^=ax​i^+ay​j^​+az​k^

Is instantaneous velocity mathematically defined as dr⃗/dtd\vec{r}/dtdr/dt?

Conditions for Straight and Curved Paths

The relationship between the velocity v⃗\vec{v}v and acceleration a⃗\vec{a}a vectors determines the shape of the object's path.

Condition for Straight-Line Motion

An object moves in a straight line if its velocity and acceleration vectors are parallel (or anti-parallel). This means the acceleration vector can only change the magnitude of the velocity (the speed), not its direction.

Mathematically, a⃗\vec{a}a is parallel to v⃗\vec{v}v. This includes the case where a⃗=0\vec{a} = 0a=0 (constant velocity).

For straight-line motion, acceleration is parallel to velocity.
For straight-line motion, acceleration is parallel to velocity.

Condition for Curved Motion

An object moves along a curved path if its acceleration vector has a component perpendicular to its velocity vector. This perpendicular component changes the direction of the velocity, causing the path to curve.

For curved motion, acceleration has a component perpendicular to velocity.
For curved motion, acceleration has a component perpendicular to velocity.

The velocity vector is always tangent to the curved path.

The velocity vector is always tangent to the path.
The velocity vector is always tangent to the path.

The acceleration vector can be resolved into two components: one parallel to the velocity (changing the speed) and one perpendicular to the velocity (changing the direction).

The acceleration vector can have components parallel and perpendicular to the velocity.
The acceleration vector can have components parallel and perpendicular to the velocity.

True or False: An object will move in a curved path if its acceleration is parallel to its velocity.

Kinematic Equations for Constant Acceleration in Vector Form

When the acceleration vector a⃗\vec{a}a is constant, the one-dimensional kinematic equations can be extended to their vector forms:

v⃗=u⃗+a⃗t\qquad \vec{v} = \vec{u} + \vec{a}tv=u+at

r⃗−r0⃗=u⃗t+12a⃗t2\qquad \vec{r} - \vec{r_0} = \vec{u}t + \frac{1}{2}\vec{a}t^2r−r0​​=ut+21​at2

where u⃗\vec{u}u is the initial velocity vector and r0⃗\vec{r_0}r0​​ is the initial position vector.

Component-wise Equations

The power of the vector form is that it can be broken down into a set of independent equations for each coordinate axis (x, y, and z). This means we can analyze the motion along each axis separately.

For the x-direction:

vx=ux+axt\qquad v_x = u_x + a_x tvx​=ux​+ax​t

x−x0=uxt+12axt2\qquad x - x_0 = u_x t + \frac{1}{2}a_x t^2x−x0​=ux​t+21​ax​t2

For the y-direction:

vy=uy+ayt\qquad v_y = u_y + a_y tvy​=uy​+ay​t

y−y0=uyt+12ayt2\qquad y - y_0 = u_y t + \frac{1}{2}a_y t^2y−y0​=uy​t+21​ay​t2

For the z-direction (if in 3D):

vz=uz+azt\qquad v_z = u_z + a_z tvz​=uz​+az​t

z−z0=uzt+12azt2\qquad z - z_0 = u_z t + \frac{1}{2}a_z t^2z−z0​=uz​t+21​az​t2

This principle of treating components independently is fundamental to solving problems in multi-dimensional kinematics, such as projectile motion.

True or False: Using component form allows us to treat motion in each direction (x, y, z) independently.

A car drives at a steady speed around a circular track. Is it accelerating?

A particle's acceleration is antiparallel to its velocity, the two arrows point in exactly opposite directions. What is its path?

02

Ground-to-Ground Projectile Motion

Introduction to Projectile Motion

Projectile motion is a fundamental concept in physics that describes the motion of an object projected into the air, subject only to the acceleration of gravity. This two-dimensional motion is seen in thrown balls, launched rockets, and fired bullets.

Assumptions for Ideal Projectile Motion:

  • Air resistance is neglected.
  • The acceleration due to gravity (ggg) is constant and acts vertically downwards.

Under these assumptions, projectile motion is confined to a vertical plane.

The parabolic trajectory of a projectile launched at an angle theta
The parabolic trajectory of a projectile launched at an angle θ\thetaθ

True or False: In projectile motion, the primary force acting on the object after launch is air resistance.

Key Terminology

  • Trajectory: The path followed by the projectile (a parabola in ideal conditions).
  • Angle of Projection (θ\thetaθ): The angle between the initial velocity vector and the horizontal.
  • Initial Velocity (uuu): The velocity with which the projectile is launched.
  • Range (R): The horizontal distance covered by the projectile.
  • Maximum Height (H): The maximum vertical displacement reached.
  • Time of Flight (T): The total time the projectile is in the air.

All six live on one picture. Five of them you can point at; the sixth is the clock.

Six words, one trajectory. Five of them are distances or angles you can point at; the sixth is a clock reading.
Six words, one trajectory. Five of them are distances or angles you can point at; the sixth is a clock reading.

A ball is launched at 55∘55^\circ55∘, traces a curved path 90 m90\,\text{m}90m long through the air, and comes down 70 m70\,\text{m}70m from the point where it left the ground. What is its range?

Concept Check: two identical balls leave a table at the same instant, one is simply nudged off the edge, the other is fired horizontally at high speed. Which one hits the floor first?

Think about the answer first, then use the interactive controls to verify.

Both balls are released at the same instant from the same height. The horizontal speed of the second one changes where it lands, not when.
Both balls are released at the same instant from the same height. The horizontal speed of the second one changes where it lands, not when.

Both balls remain at the exact same height throughout the fall and hit the ground at the same time. This demonstrates the principle of independence of vertical and horizontal motions: gravity acts only vertically downwards and does not affect the horizontal velocity. Therefore, the horizontal motion does not affect the vertical free fall. The vertical motion does not affect the horizontal motion.

All calculations for projectile motion are based on this principle. The horizontal and vertical components of motion are independent and share only the time variable ttt.

Independence of Horizontal and Vertical Motion

The core principle for analyzing projectile motion is the physical independence of the horizontal and vertical components of motion. Since gravity acts only vertically, we can treat the two components separately:

  • Horizontal Motion: There is no acceleration in the horizontal direction. Consequently, the horizontal velocity is constant (uniform motion).
  • Vertical Motion: There is a constant downward acceleration ggg, so the vertical motion is uniformly accelerated motion.

Resolving Initial Velocity

If a projectile is launched with an initial velocity uuu at an angle θ\thetaθ to the horizontal, we resolve this vector into its components:

ux=ucos⁡θ(Horizontal component)\qquad u_x = u\cos\theta \quad \text{(Horizontal component)}ux​=ucosθ(Horizontal component)

uy=usin⁡θ(Vertical component)\qquad u_y = u\sin\theta \quad \text{(Vertical component)}uy​=usinθ(Vertical component)

The initial velocity vector is resolved into its x and y components.
The initial velocity vector is resolved into its x and y components.

The two shadows

Concept Check: a lamp overhead casts the ball's shadow on the floor. A lamp shining sideways casts its shadow on the wall. Throw the ball and watch the two shadows. How does each one move?

Think about the answer first, then use the interactive controls to verify.

A lamp overhead prints the ball's shadow on the floor; a lamp at the side prints it on the far wall. Drag to orbit, scroll to zoom, and use the controls to throw the ball.

The shadow on the floor moves at a steady speed. It never speeds up and it never slows down, because nothing is pushing the ball sideways. The shadow on the wall does something quite different. It rises, slows down, stops for an instant at the top. Then falls back faster and faster, exactly like a ball thrown straight up.

The real ball is doing both of these at the same time. The floor shadow is its horizontal motion and the wall shadow is its vertical motion. The only thing the two share is the clock.

One flight at equal time intervals. The floor shadow advances evenly; the wall shadow slows, stops, and returns.
One flight at equal time intervals. The floor shadow advances evenly; the wall shadow slows, stops, and returns.

Mark the flight at equal intervals of time and the same story shows up as spacing: the floor marks are evenly spaced, while the wall marks crowd together near the top. Pick any instant, read both marks, and you have the position of the ball. That is what independence means, and it is also the method: solve two independent 1D motion problems and link them using time ttt.

True or False: In projectile motion, the horizontal and vertical components of motion are physically independent.

Kinematic Equations for Projectile Motion

By applying the standard kinematic equations to each component separately:

Horizontal Motion (Constant Velocity)

vx=ux=ucos⁡θ\qquad v_x = u_x = u\cos\thetavx​=ux​=ucosθ

x=uxt=(ucos⁡θ)t\qquad x = u_x t = (u\cos\theta)tx=ux​t=(ucosθ)t

Vertical Motion (Constant Acceleration −g-g−g)

vy=uy−gt=usin⁡θ−gt\qquad v_y = u_y - gt = u\sin\theta - gtvy​=uy​−gt=usinθ−gt

y=uyt−12gt2=(usin⁡θ)t−12gt2\qquad y = u_y t - \frac{1}{2}gt^2 = (u\sin\theta)t - \frac{1}{2}gt^2y=uy​t−21​gt2=(usinθ)t−21​gt2

vy2=uy2−2gy=(usin⁡θ)2−2gy\qquad v_y^2 = u_y^2 - 2gy = (u\sin\theta)^2 - 2gyvy2​=uy2​−2gy=(usinθ)2−2gy

Two one-dimensional problems side by side. The only thing shared between the columns is the clock.
Two one-dimensional problems side by side. The only thing shared between the columns is the clock.

Time of Flight, Maximum Height, and Range

Time of Flight (T)

The time of flight is the total time the projectile is in the air. For a projectile that lands at the same height it was launched from, the vertical displacement yyy is zero. We use the vertical motion equation:

0=(usin⁡θ)T−12gT20 = (u\sin\theta)T - \frac{1}{2}gT^20=(usinθ)T−21​gT2

Solving for TTT (for T>0T > 0T>0):

T=2usin⁡θg=2uyg\qquad T = \dfrac{2u\sin\theta}{g} = \dfrac{2u_y}{g}T=g2usinθ​=g2uy​​

The second form is the same answer written with the components. The vertical component decides the time of flight on its own; uxu_xux​ does not appear at all.

Maximum Height (H)

At the maximum height of the trajectory, the vertical component of the velocity vyv_yvy​ is momentarily zero. Using the vertical velocity equation:

02=(usin⁡θ)2−2gH0^2 = (u\sin\theta)^2 - 2gH02=(usinθ)2−2gH

Solving for HHH:

H=u2sin⁡2θ2g=uy22g\qquad H = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{u_y^{2}}{2g}H=2gu2sin2θ​=2guy2​​

In components, again the vertical one alone: two balls with the same uyu_yuy​ rise equally high, whatever their sideways speed.

Horizontal Range (R)

The range is the horizontal distance traveled during the time of flight. Using the horizontal motion equation:

R=uxT=(ucos⁡θ)(2usin⁡θg)=u2(2sin⁡θcos⁡θ)gR = u_x T = (u\cos\theta) \left( \dfrac{2u\sin\theta}{g} \right) = \dfrac{u^2(2\sin\theta\cos\theta)}{g}R=ux​T=(ucosθ)(g2usinθ​)=gu2(2sinθcosθ)​

Using the trigonometric identity sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\thetasin(2θ)=2sinθcosθ:

R=u2sin⁡(2θ)g\qquad R = \dfrac{u^2\sin(2\theta)}{g}R=gu2sin(2θ)​

In components this is just the first line again:

R=uxT=2uxuyg\qquad R = u_x T = \dfrac{2u_x u_y}{g}R=ux​T=g2ux​uy​​

Range is the only one of the three that uses uxu_xux​. Only as a steady speed kept up for the time TTT the vertical motion has already fixed.

The range is maximum when sin⁡(2θ)=1\sin(2\theta) = 1sin(2θ)=1, which occurs at 2θ=90∘2\theta = 90^\circ2θ=90∘ or θ=45∘\theta = 45^\circθ=45∘.

The three together. TTT and HHH depend on uyu_yuy​ alone, while R=uxTR = u_x TR=ux​T. Change uxu_xux​ and you change only where the ball lands.

Where each result actually comes from

Concept Check: three formulas, three derivations. Which of TTT, HHH and RRR could you find without ever looking at the horizontal motion at all?

Think about the answer first, then use the interactive controls to verify.

Each tab highlights the part of the flight that produces that one result.
Each tab highlights the part of the flight that produces that one result.

Two of the three come from the vertical motion alone. For the time of flight TTT we ask when the ball is back at its starting level; for the maximum height HHH we ask when its upward speed falls to zero. Both are questions about up and down only. Consequently, both answers contain the vertical component and nothing else:

T=2uyg,H=uy22g\qquad T = \dfrac{2u_y}{g}, \qquad H = \dfrac{u_y^{2}}{2g}T=g2uy​​,H=2guy2​​

The range is the only one that needs the sideways motion. The ball moves sideways at a steady speed uxu_xux​ for the time TTT the vertical motion has already fixed, so R=uxT=2uxuygR = u_x T = \dfrac{2u_x u_y}{g}R=ux​T=g2ux​uy​​. Easier to remember as a sentence: range is sideways speed times the time the vertical motion allows.

Three launches with the same vertical component uy but different horizontal components ux. All three reach the same height and stay up for the same time; only the landing point moves.
Three launches with the same vertical component uyu_yuy​ but different horizontal components uxu_xux​. All three reach the same height and stay up for the same time; only the landing point moves.

Keep uyu_yuy​ fixed and change uxu_xux​, as in the figure: the height and the time in the air do not move at all, while the landing point slides further out. This illustrates the independence of horizontal and vertical motions.

Equation of Trajectory

By eliminating the time variable ttt from the horizontal and vertical position equations, we can find the equation of the projectile's path (its trajectory):

y=xtan⁡θ−gx22u2cos⁡2θ\qquad y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}y=xtanθ−2u2cos2θgx2​

This is the equation of a parabola opening downwards, confirming that the trajectory of a projectile is parabolic. The origin is the launch point, with yyy measured upwards from there.

An alternative form of this equation is:

y=xtan⁡θ(1−xR)\qquad y = x\tan\theta \left(1 - \dfrac{x}{R}\right)y=xtanθ(1−Rx​)

Level ground only. This second form works only when the ball lands at the height it was launched from. This occurs because the RRR in it was found by setting y=0y=0y=0 at landing. The first form has no such limit.

The parabolic trajectory of a projectile, showing the relationship between x and y.
The parabolic trajectory of a projectile, showing the relationship between x and y.

Velocity Components During Flight

The velocity vector changes continuously throughout the flight. The horizontal component vxv_xvx​ remains constant, while the vertical component vyv_yvy​ changes due to gravity.

The velocity vector and its components are shown at different points along the trajectory.
The velocity vector and its components are shown at different points along the trajectory.

Is the trajectory of a projectile in ideal projectile motion described as a parabolic path?

Problem 1: Basic Projectile Motion

A ball is thrown with a velocity of 20 m/s at an angle of 30° above the horizontal. Find: (a) the maximum height, (b) the time of flight, and (c) the horizontal range. (Use g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2)

The situation before any algebra: the height and the range are still to be found.
The situation before any algebra: the height and the range are still to be found.
Show the solution

(a) Maximum height:

H=u2sin⁡2θ2g=(20)2sin⁡2(30∘)2(10)=400×(0.5)220=5 mH = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{(20)^2 \sin^2(30^\circ)}{2(10)} = \dfrac{400 \times (0.5)^2}{20} = 5 \, \text{m}H=2gu2sin2θ​=2(10)(20)2sin2(30∘)​=20400×(0.5)2​=5m

(b) Time of flight:

T=2usin⁡θg=2(20)sin⁡(30∘)10=40×0.510=2 sT = \dfrac{2u\sin\theta}{g} = \dfrac{2(20)\sin(30^\circ)}{10} = \dfrac{40 \times 0.5}{10} = 2 \, \text{s}T=g2usinθ​=102(20)sin(30∘)​=1040×0.5​=2s

(c) Range:

R=u2sin⁡(2θ)g=(20)2sin⁡(60∘)10=400×(3/2)10=203≈34.6 mR = \dfrac{u^2\sin(2\theta)}{g} = \dfrac{(20)^2\sin(60^\circ)}{10} = \dfrac{400 \times (\sqrt{3}/2)}{10} = 20\sqrt{3} \approx 34.6 \, \text{m}R=gu2sin(2θ)​=10(20)2sin(60∘)​=10400×(3​/2)​=203​≈34.6m

A ball rolls off a table at 3 m/s3\,\text{m/s}3m/s. A second ball is dropped from the table edge at the same instant. Which reaches the floor first?

A projectile launched at 30∘30^\circ30∘ has a range of 35 m35\,\text{m}35m. Fired at the same speed at 60∘60^\circ60∘, what is its range?

Exploring the launch angle

Concept Check: you can fire at any angle you like with a fixed speed. Which angle sends the projectile furthest. Is there more than one angle that reaches a given distance?

Think about the answer first, then use the interactive controls to verify.

The same launch speed at five angles. 45° reaches furthest; angles that add to 90° land in the same place by very different routes.
The same launch speed at five angles. 45° reaches furthest; angles that add to 90° land in the same place by very different routes.

Two results are worth taking away from this. The range peaks at 45∘45^\circ45∘, because R=u2sin⁡2θgR = \dfrac{u^2\sin 2\theta}{g}R=gu2sin2θ​ and sin⁡2θ\sin 2\thetasin2θ is largest when 2θ=90∘2\theta = 90^\circ2θ=90∘. And complementary angles share a range: 30∘30^\circ30∘ and 60∘60^\circ60∘ land together, as do 15∘15^\circ15∘ and 75∘75^\circ75∘. This occurs because sin⁡2θ\sin 2\thetasin2θ takes the same value at θ\thetaθ and at 90∘−θ90^\circ-\theta90∘−θ.

They get there completely differently, though. Watch the time of flight and the maximum height as you switch between a pair: the steeper shot trades horizontal speed for airtime, and the trade is exactly even.

The same result, as a curve

The same result seen a different way, not the paths through space, but the single number each path delivers:

Range against launch angle. One arch, symmetric about 45°, which is the complementary-angle rule made visible.
Range against launch angle. One arch, symmetric about 45°, which is the complementary-angle rule made visible.

In this picture the complementary-angle result stops being a trigonometric coincidence and becomes a statement about a shape: the curve is symmetric about 45∘45^\circ45∘. Consequently, reading it at θ\thetaθ and at 90∘−θ90^\circ - \theta90∘−θ gives the same height. It also makes the flatness near the peak visible, anywhere between about 40∘40^\circ40∘ and 50∘50^\circ50∘ the range is within 2% of its maximum. This is why aiming is forgiving near 45° and unforgiving near 0° or 90°.

Two checks worth making every time

Check it at the edges. Send the angle to its extremes. At θ=0\theta = 0θ=0 the formulas give T=0T = 0T=0, H=0H = 0H=0, R=0R = 0R=0, fire horizontally from ground level and the projectile is already on the ground, which is right. At θ=90∘\theta = 90^\circθ=90∘, R=0R = 0R=0 but H=u2/2gH = u^2/2gH=u2/2g is largest, straight up and straight back down. Any formula you write for range must vanish at both ends and peak in the middle; if yours does not, it is wrong before you put numbers in.

Example: the wrong answer that feels right

A student is asked for the speed of a projectile at the top of its flight, launched at 20 m/s20\,\text{m/s}20m/s at 30∘30^\circ30∘. They answer: "At the top it stops momentarily, so the speed is zero."

Where exactly does that reasoning fail. What is the correct answer?

Show the solution

What is right about it. Something does become zero at the top, the vertical component vyv_yvy​. The student has correctly identified the defining property of the highest point.

Where it fails. The reasoning quietly imports a one-dimensional habit: in 1D, "velocity is zero" and "momentarily at rest" mean the same thing. In 2D they do not. Only one component vanishes; the other is untouched, because gravity never acted on it.

The correct answer. vx=ucos⁡θ=20cos⁡30°=17.3 m/sv_x = u\cos\theta = 20\cos 30° = 17.3\,\text{m/s}vx​=ucosθ=20cos30°=17.3m/s, constant for the whole flight. At the top vy=0v_y = 0vy​=0, so the speed there is 17.3 m/s17.3\,\text{m/s}17.3m/s, horizontal. It is the minimum speed of the flight, but it is not zero.

The general lesson. A projectile is only at rest if it is fired straight up. Any launch with a horizontal component keeps that component forever.

What stays the same, and what does not

The blue arrow is the same length at every point of the flight. Gravity acts straight down, so the green one is the only thing it can touch.
The blue arrow is the same length at every point of the flight. Gravity acts straight down, so the green one is the only thing it can touch.

This is the independence principle drawn on a single trajectory. The horizontal component is untouched from launch to landing. The vertical component shrinks, passes through zero at the top. Then grows downward at the same steady rate it was shrinking, which is why the flight is symmetric about the highest point.

Two projectiles are launched with different speeds and different angles. Both trajectories are parabolas. What does that tell you?

Problem 3: Finding Initial Velocity

A projectile is launched at an angle of 45° with the horizontal and has a range of 100 m. Find the initial velocity. (Use g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2)

Same picture, different unknown, the range is given, the speed is not.
Same picture, different unknown, the range is given, the speed is not.
Show the solution

Given: R=100 mR = 100 \, \text{m}R=100m, θ=45∘\theta = 45^\circθ=45∘, g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2

Since R=u2sin⁡2θgR = \dfrac{u^2\sin2\theta}{g}R=gu2sin2θ​, we have 100=u2sin⁡(90∘)10100 = \dfrac{u^2\sin(90^\circ)}{10}100=10u2sin(90∘)​

100=u210100 = \dfrac{u^2}{10}100=10u2​

u2=1000u^2 = 1000u2=1000

u=1000=1010 m/su = \sqrt{1000} = 10\sqrt{10} \, \text{m/s}u=1000​=1010​m/s

Problem 4: Finding Launch Angle

A projectile is launched with an initial velocity of 202 m/s20\sqrt{2} \, \text{m/s}202​m/s. If the range of the projectile is 80 m, find the launch angle. (Use g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2)

Speed and range given, angle unknown. Complementary angles share a range.
Speed and range given, angle unknown. Complementary angles share a range.
Show the solution

Given: u=202 m/su = 20\sqrt{2} \, \text{m/s}u=202​m/s, R=80 mR = 80 \, \text{m}R=80m, g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2

Since R=u2sin⁡2θgR = \dfrac{u^2\sin2\theta}{g}R=gu2sin2θ​, we have 80=(202)2sin⁡2θ1080 = \dfrac{(20\sqrt{2})^2\sin2\theta}{10}80=10(202​)2sin2θ​

80=800sin⁡2θ1080 = \dfrac{800\sin2\theta}{10}80=10800sin2θ​

sin⁡2θ=1\sin2\theta = 1sin2θ=1

2θ=90∘2\theta = 90^\circ2θ=90∘

θ=45∘\theta = 45^\circθ=45∘

Now check for a second angle. A given range usually has two launch angles. This occurs because sin⁡2θ\sin 2\thetasin2θ takes the same value at 2θ2\theta2θ and at 180∘−2θ180^\circ - 2\theta180∘−2θ. Here the complement of 45∘45^\circ45∘ is 45∘45^\circ45∘ itself, so this is the one case with a single answer: 80 m80\ \text{m}80 m is the farthest this speed can reach, because Rmax⁡=u2/g=800/10=80 mR_{\max} = u^2/g = 800/10 = 80\ \text{m}Rmax​=u2/g=800/10=80 m. There is exactly one way to just reach it.

Ask for a shorter range and the pair appears. For R=70 mR = 70\ \text{m}R=70 m, sin⁡2θ=0.875\sin 2\theta = 0.875sin2θ=0.875, giving 2θ=61.0∘2\theta = 61.0^\circ2θ=61.0∘ or 119.0∘119.0^\circ119.0∘ — that is θ=30.5∘\theta = 30.5^\circθ=30.5∘ or 59.5∘59.5^\circ59.5∘, two angles adding to 90∘90^\circ90∘.

03

Projectile Motion from a Height

Projectile Motion from a Height

When a projectile is launched from an elevated position, such as a cliff or a building, the initial and final vertical positions are different. This introduces a non-zero vertical displacement. This affects the time of flight and, consequently, the horizontal range.

The same principles of independent horizontal and vertical motion apply, but the vertical displacement equation must be solved as a quadratic equation to find the time of flight.

A particle projected from the top of a building.
A particle projected from the top of a building.

Aim up, level, or down?

Concept Check: you can fire the same speed off the top of a cliff aimed upwards, level, or downwards. Which shot stays in the air longest. Which lands furthest from the cliff?

Think about the answer first, then use the interactive controls to verify.

One speed, one cliff, three aims. Aiming down lands soonest and nearest; aiming up does the opposite.
One speed, one cliff, three aims. Aiming down lands soonest and nearest; aiming up does the opposite.

When a projectile is launched from a height hhh above the ground, the vertical displacement at landing is y=−hy = -hy=−h instead of y=0y = 0y=0. Substituting y=−hy = -hy=−h into the position equation y=(usin⁡θ)t−12gt2y = (u\sin\theta)t - \tfrac12 gt^{2}y=(usinθ)t−21​gt2 yields a quadratic equation in time ttt. Solving this quadratic equation gives two roots. The positive root represents the actual time of flight. The negative root corresponds to the time at which a ball thrown from ground level would have passed the cliff top at t=0t=0t=0 with the same velocity.

Important Note: The standard formulas for time of flight T=2usin⁡θgT = \dfrac{2u\sin\theta}{g}T=g2usinθ​ and horizontal range R=u2sin⁡2θgR = \dfrac{u^{2}\sin 2\theta}{g}R=gu2sin2θ​ cannot be used here because they were derived for launch and landing at the same level (y=0y = 0y=0). When the launch and landing heights are different (y=−hy = -hy=−h), always use the individual motion equations along the x and y axes.

On level ground, launches at 30∘30^\circ30∘ and 60∘60^\circ60∘ with the same speed land the same distance away. Fired at that same speed from the top of a cliff, how do their distances from the base of the cliff compare?

Example Problems

Problem 2: Launch from a Cliff

A ball is thrown with a velocity of 20 m/s at an angle of 30° above the horizontal from the top of a 40 m high cliff. Find the time it takes to hit the ground and the horizontal range.

The same launch, now from 40 m up: the landing point is below the launch point.
The same launch, now from 40 m up: the landing point is below the launch point.
Show the solution

Initial velocity components: ux=20cos⁡(30∘)=103 m/su_x = 20\cos(30^\circ) = 10\sqrt{3} \, \text{m/s}ux​=20cos(30∘)=103​m/s, uy=20sin⁡(30∘)=10 m/su_y = 20\sin(30^\circ) = 10 \, \text{m/s}uy​=20sin(30∘)=10m/s.

To find the time, we use the vertical motion equation with a displacement of y=−40 my = -40 \, \text{m}y=−40m:

−40=10t−12(10)t2⇒5t2−10t−40=0-40 = 10t - \frac{1}{2}(10)t^2 \Rightarrow 5t^2 - 10t - 40 = 0−40=10t−21​(10)t2⇒5t2−10t−40=0

t2−2t−8=0⇒(t−4)(t+2)=0t^2 - 2t - 8 = 0 \Rightarrow (t-4)(t+2) = 0t2−2t−8=0⇒(t−4)(t+2)=0

The time of flight is t=4 st = 4 \, \text{s}t=4s.

The range is x=uxt=(103)(4)=403≈69.3 mx = u_x t = (10\sqrt{3})(4) = 40\sqrt{3} \approx 69.3 \, \text{m}x=ux​t=(103​)(4)=403​≈69.3m.

04

The Equation of Trajectory

Why Eliminate Time from Motion Equations?

In projectile motion problems, we often need to find the position of a projectile in space, such as its height at a specific horizontal distance or whether it clears an obstacle, without needing to know the time taken. Calculating time first is an indirect step when time is not required by the problem.

The equation of trajectory provides a direct mathematical relationship between the vertical position yyy and horizontal distance xxx by eliminating time ttt. This equation allows us to find the vertical height at any horizontal distance directly.

To eliminate time ttt, we express time from the horizontal motion equation x=(ucos⁡θ)tx = (u\cos\theta)tx=(ucosθ)t as t=xucos⁡θt = \dfrac{x}{u\cos\theta}t=ucosθx​. Substituting this expression for ttt into the vertical displacement equation y=(usin⁡θ)t−12gt2y = (u\sin\theta)t - \frac{1}{2}gt^2y=(usinθ)t−21​gt2 gives the direct height formula y=xtan⁡θ−gx22u2cos⁡2θy = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta}y=xtanθ−2u2cos2θgx2​. This allows us to calculate the exact vertical height yyy at any given horizontal position xxx without calculating time.

A ball is thrown from ground level. You need to know whether it clears a 3 m3\,\text{m}3m wall standing 12 m12\,\text{m}12m away. Which relation answers that most directly?

Example: Finding Height at a Given Horizontal Distance (Clearing a Wall)

Finding the vertical height of a projectile at a specific horizontal distance.
Finding the vertical height of a projectile at a specific horizontal distance.

To find the vertical height yyy of a projectile at a given horizontal distance xxx, substitute the value of xxx directly into the trajectory equation:

y=xtan⁡θ−gx22u2cos⁡2θy = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta}y=xtanθ−2u2cos2θgx2​

Alternatively, you can first calculate the time taken to travel horizontal distance xxx:

t=xucos⁡θt = \frac{x}{u\cos\theta}t=ucosθx​

and then substitute ttt into the vertical displacement formula y=(usin⁡θ)t−12gt2y = (u\sin\theta)t - \frac{1}{2}gt^2y=(usinθ)t−21​gt2. Both methods yield the exact same result. The trajectory equation is much faster when time is not asked in the question.

Concept Check: Given x=(ucos⁡θ)tx = (u\cos\theta)tx=(ucosθ)t and y=(usin⁡θ)t−12gt2y = (u\sin\theta)t - \frac{1}{2}gt^2y=(usinθ)t−21​gt2, if you eliminate time ttt between these two equations, what mathematical shape does y(x)y(x)y(x) represent, a straight line, a parabola, or an ellipse?

Think about the answer first, then use the interactive controls to verify.

Derivation steps: Eliminating time t yields a quadratic relationship between y and x with a negative coefficient of x².
Derivation steps: Eliminating time t yields a quadratic relationship between y and x with a negative coefficient of x².

Derivation of Equation of Trajectory

Since horizontal velocity is constant (ax=0a_x = 0ax​=0), horizontal displacement xxx after time ttt is given by:

x=(ucos⁡θ) t⇒t=xucos⁡θx = (u\cos\theta)\,t \qquad\Rightarrow\qquad t = \frac{x}{u\cos\theta}x=(ucosθ)t⇒t=ucosθx​

Since horizontal speed is constant, horizontal distance xxx is directly proportional to time ttt. Now, substitute this expression for ttt into the vertical displacement equation y=(usin⁡θ)t−12gt2y = (u\sin\theta)t - \frac{1}{2}gt^2y=(usinθ)t−21​gt2:

y=(usin⁡θ)(xucos⁡θ)−g2(xucos⁡θ)2y = (u\sin\theta)\left(\frac{x}{u\cos\theta}\right) - \frac{g}{2}\left(\frac{x}{u\cos\theta}\right)^{2}y=(usinθ)(ucosθx​)−2g​(ucosθx​)2

Using sin⁡θcos⁡θ=tan⁡θ\dfrac{\sin\theta}{\cos\theta} = \tan\thetacosθsinθ​=tanθ, we get the standard equation of trajectory:

  y=xtan⁡θ−g x22u2cos⁡2θ  \boxed{\; y = x\tan\theta - \frac{g\,x^{2}}{2u^{2}\cos^{2}\theta} \;}y=xtanθ−2u2cos2θgx2​​

Important Conventions: The launch point is taken as the origin (0,0)(0,0)(0,0). Horizontal distance xxx is measured along the ground, and vertical height yyy is measured upwards from the launch level. If the projectile falls below the launch level, yyy becomes negative.

Relationship between horizontal distance x and time t along a constant-velocity axis.
Relationship between horizontal distance x and time t along a constant-velocity axis.

Since horizontal velocity is uniform, every equal interval of horizontal distance corresponds to an equal interval of time.

In the derivation, why is it the horizontal equation that gets solved for ttt, rather than the vertical one?

Mathematical Proof: Why the Path is Always a Parabola

For a given initial launch speed uuu and angle θ\thetaθ, the terms tan⁡θ\tan\thetatanθ and g2u2cos⁡2θ\dfrac{g}{2u^2\cos^2\theta}2u2cos2θg​ are constants. Let A=tan⁡θA = \tan\thetaA=tanθ and B=g2u2cos⁡2θB = \dfrac{g}{2u^2\cos^2\theta}B=2u2cos2θg​. The trajectory equation simplifies to:

y=Ax−Bx2(A>0,  B>0)y = Ax - Bx^{2} \qquad (A > 0,\; B > 0)y=Ax−Bx2(A>0,B>0)

This is a quadratic equation in xxx with a negative coefficient of x2x^2x2. In mathematics, any equation of the form y=Ax−Bx2y = Ax - Bx^2y=Ax−Bx2 represents an inverted (downward-opening) parabola.

Physical Explanation: Horizontal displacement xxx is linear with respect to time (x∝tx \propto tx∝t), while vertical displacement yyy is quadratic with respect to time (y∝t2y \propto t^2y∝t2). Eliminating ttt yields y∝x2y \propto x^2y∝x2. Whenever one axis has uniform motion and the perpendicular axis has uniform acceleration, the resulting path is always a parabola.

Combination of uniform horizontal motion and uniformly accelerated vertical motion forms a parabolic trajectory.
Combination of uniform horizontal motion and uniformly accelerated vertical motion forms a parabolic trajectory.

Effect of Launch Parameters on Trajectory

Concept Check: If you vary the launch speed uuu and launch angle θ\thetaθ, does the path ever stop being a downward-opening parabola?

Think about the answer first, then use the interactive controls to verify.

Trajectories for different launch angles. The straight dashed line represents motion without gravity.
Trajectories for different launch angles. The straight dashed line represents motion without gravity.

The straight dashed line y=Axy = Axy=Ax represents the hypothetical straight-line path of the projectile if gravity were zero (g=0g = 0g=0). The vertical distance dropped below this straight line is Bx2=12gt2Bx^2 = \frac{1}{2}gt^2Bx2=21​gt2, which represents the effect of gravity pulling the projectile downwards.

Factored Form of Trajectory Equation

Factoring out xtan⁡θx\tan\thetaxtanθ from the standard trajectory equation gives:

y=xtan⁡θ(1−g x2u2sin⁡θcos⁡θ)y = x\tan\theta\left(1 - \frac{g\,x}{2u^{2}\sin\theta\cos\theta}\right)y=xtanθ(1−2u2sinθcosθgx​)

Since horizontal range on level ground is R=u2sin⁡2θg=2u2sin⁡θcos⁡θgR = \dfrac{u^2\sin 2\theta}{g} = \dfrac{2u^2\sin\theta\cos\theta}{g}R=gu2sin2θ​=g2u2sinθcosθ​, substituting RRR into the equation yields the factored form of the trajectory equation:

  y=xtan⁡θ(1−xR)  \boxed{\; y = x\tan\theta\left(1 - \frac{x}{R}\right) \;}y=xtanθ(1−Rx​)​

Key Properties of the Factored Form:

  • At x=0x = 0x=0 and x=Rx = Rx=R, the vertical displacement is y=0y = 0y=0 (launch point and landing point).
  • Because a parabola is symmetric, the maximum height occurs exactly halfway along the range, at x=R2x = \dfrac{R}{2}x=2R​.
  • Important Limitation: This factored form applies only to ground-to-ground projectile motion, where landing height equals launch height.
Factored form of trajectory equation showing roots at x = 0 and x = R, with maximum height at x = R/2.
Factored form of trajectory equation showing roots at x = 0 and x = R, with maximum height at x = R/2.

From y=xtan⁡θ(1−xR)y = x\tan\theta\left(1 - \dfrac{x}{R}\right)y=xtanθ(1−Rx​), where is the projectile highest?

Example: Finding Launch Speed and Launch Angle from Trajectory Equation

Worked Problem

The trajectory equation of a projectile is given by y=x−x240y = x - \dfrac{x^{2}}{40}y=x−40x2​, where xxx and yyy are in metres and g=10 m/s2g = 10\,\text{m/s}^{2}g=10m/s2.

Find: (a) Launch angle θ\thetaθ, (b) Initial speed uuu, (c) Horizontal range RRR. (d) Maximum height HHH.

Show the solution

Step 1: Compare with standard equation:

y=xtan⁡θ−gx22u2cos⁡2θy = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta}y=xtanθ−2u2cos2θgx2​

Step 2: Find launch angle θ\thetaθ:
Comparing the coefficient of xxx:

tan⁡θ=1⇒θ=45∘\tan\theta = 1 \quad\Rightarrow\quad \theta = 45^\circtanθ=1⇒θ=45∘

Step 3: Find launch speed uuu:
Comparing the coefficient of x2x^2x2, noting cos⁡45∘=12\cos 45^\circ = \frac{1}{\sqrt{2}}cos45∘=2​1​ and cos⁡245∘=12\cos^2 45^\circ = \frac{1}{2}cos245∘=21​:

g2u2cos⁡2θ=102u2(1/2)=10u2=140\frac{g}{2u^2\cos^2\theta} = \frac{10}{2u^2(1/2)} = \frac{10}{u^2} = \frac{1}{40}2u2cos2θg​=2u2(1/2)10​=u210​=401​u2=400⇒u=20 m/su^2 = 400 \quad\Rightarrow\quad u = 20\ \text{m/s}u2=400⇒u=20 m/s

Step 4: Find Horizontal Range RRR:
Rewriting the given equation in factored form y=x(1−x40)y = x\left(1 - \frac{x}{40}\right)y=x(1−40x​), comparing with y=xtan⁡θ(1−xR)y = x\tan\theta\left(1 - \frac{x}{R}\right)y=xtanθ(1−Rx​) gives:

R=40 mR = 40\ \text{m}R=40 m

Alternatively, using R=u2sin⁡2θg=400×sin⁡90∘10=40 mR = \frac{u^2\sin 2\theta}{g} = \frac{400 \times \sin 90^\circ}{10} = 40\ \text{m}R=gu2sin2θ​=10400×sin90∘​=40 m.

Step 5: Find Maximum Height HHH:
At maximum height, x=R2=20 mx = \frac{R}{2} = 20\ \text{m}x=2R​=20 m:

ymax⁡=20(1−2040)=10 my_{\max} = 20\left(1 - \frac{20}{40}\right) = 10\ \text{m}ymax​=20(1−4020​)=10 m
Verification of initial speed u, launch angle θ, range R, and maximum height H from trajectory equation.
Verification of initial speed u, launch angle θ, range R, and maximum height H from trajectory equation.
Comparing coefficients of x and x² to find launch parameters.
Comparing coefficients of x and x² to find launch parameters.

A projectile's path is y=2x−x220y = 2x - \dfrac{x^{2}}{20}y=2x−20x2​. What is the launch angle?

Limitations and Validity of Parabolic Trajectory

The parabolic trajectory equation is derived under three key assumptions:

  1. Acceleration due to gravity ggg is constant in magnitude.
  2. Gravity acts vertically downwards everywhere.
  3. Air resistance (drag) is completely negligible.

If any of these conditions are violated, the trajectory deviates from a true parabola:

  • Air Resistance: Drag increases with speed, causing the descending path to be steeper and shorter than the ascending path. The trajectory loses its symmetry.
  • Long-Distance Motion: Over large distances across the Earth, gravity changes direction towards the Earth's center, making the trajectory elliptical rather than parabolic.
  • Earth's Rotation: Over long flight times, the Coriolis effect causes lateral deflection out of the vertical plane.

Checking Edge Cases

Checking Edge Cases: Consider the trajectory equation at extreme limits:
1. When θ=0∘\theta = 0^\circθ=0∘ (horizontal launch), tan⁡0∘=0\tan 0^\circ = 0tan0∘=0, giving y=−gx22u2y = -\frac{gx^2}{2u^2}y=−2u2gx2​. The projectile only moves downwards from the launch level.
2. As launch speed u→∞u \to \inftyu→∞, the quadratic term approaches zero (gx22u2cos⁡2θ→0\frac{gx^2}{2u^2\cos^2\theta} \to 02u2cos2θgx2​→0), approaching the straight line y=xtan⁡θy = x\tan\thetay=xtanθ.

Comparison of ideal parabolic trajectory versus real path with air resistance showing asymmetric descent.
Comparison of ideal parabolic trajectory versus real path with air resistance showing asymmetric descent.

Observe the shape of the path with air resistance. The path is still curved, but it is no longer symmetric. A true parabola is always symmetric. This asymmetry occurs because air resistance changes the acceleration of the projectile during its flight.

Limiting cases of the trajectory equation: horizontal launch and infinite speed limit.
Limiting cases of the trajectory equation: horizontal launch and infinite speed limit.

Which single change would stop a projectile's path being a parabola?

A projectile is launched from ground level and lands back at ground level, but its measured path is not symmetric about its highest point, the descent is visibly steeper than the climb. What does that tell you?

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