Projectile motion: practice problems with step-by-step solutions
Also called: projectile range
Free Projectile motion practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: horizontal launch, launch from ground, launch from height.
Free practice problemEasy
A stone is thrown horizontally from the top of a 20 m high wall and lands 22 m from the foot of the wall. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A projectile is an object that flies through the air with only gravity acting on it. A kicked ball, a stone thrown from a bridge and water from a hose are all projectiles.
In this topic you ignore air resistance. Then the path is always a smooth curve called a parabola.
The key idea: the motion splits into two separate parts. Sideways, nothing pushes or slows the object. Up and down, gravity pulls it at g = 9.81 m/s².
So the horizontal speed stays the same the whole time, while the vertical speed changes every second. The two parts share only one thing: the time.
Self-check A ball flies through the air with no air resistance. Which quantity stays constant during the flight?
The height
The horizontal velocity
The total speed
The vertical velocity
Answer: The horizontal velocity. No force acts sideways, so the horizontal velocity never changes. Gravity changes the vertical velocity, and with it the total speed and the height.
2Two motions, one clock
Put the origin at the launch point, with x horizontal and y vertical (up is positive). After a time t:
x = vx · t
y = h + vy · t − ½ · g · t²
Here h is the launch height above the ground, vx is the horizontal velocity and vy is the starting vertical velocity.
The ball hits the ground when y = 0. You solve the vertical equation for that time, then put the same time into the horizontal equation.
For a horizontal launch, vy = 0. The vertical equation becomes h = ½ · g · t², so the fall time is:
t = √(2h / g)
This fall time does not depend on the horizontal speed at all.
Self-check From the same table, ball A rolls off the edge at 1 m/s and ball B at 3 m/s. Which one lands first?
Ball A
Ball B
It depends on their masses
They land at the same time
Answer: They land at the same time. Both start with zero vertical velocity and fall the same height, so t = √(2h/g) is the same. Ball B only lands further away.
3Launching at an angle
When the ball is launched at a speed v0 and an angle θ above the horizontal, first split the velocity into its two parts, the components:
vx = v0 · cos θ vy = v0 · sin θ
From level ground, the ball rises until its vertical velocity is zero, then falls back. The rise and the fall take equal times. This gives three useful results:
ttop = vy / g t = 2 · vy / g
H = vy² / (2g) R = v0² · sin 2θ / g
The range formula for R is only valid when the ball lands at the same height it was launched from.
Self-check A ball is launched at 20 m/s and 30° above the horizontal. What are its velocity components?
vx = 10 m/s and vy = 10 m/s
vx = 10 m/s and vy = 17.3 m/s
vx = 17.3 m/s and vy = 10 m/s
vx = 20 m/s and vy = 20 m/s
Answer: vx = 17.3 m/s and vy = 10 m/s. vx = 20 · cos 30° = 17.3 m/s and vy = 20 · sin 30° = 10 m/s. Swapping them uses sin for the horizontal part, which is the classic mix-up.
4The method step by step
Draw the path. Mark the launch height h, the launch speed v0, the angle θ and what you need to find.
Split the launch velocity into components: vx = v0 · cos θ and vy = v0 · sin θ. A horizontal launch has vy = 0.
Use the vertical motion to find the time of flight: set y = 0 in y = h + vy·t − ½·g·t² and solve for t.
Use that time in the horizontal motion: x = vx · t.
For the speed at any moment, combine the components: v = √(vx² + (vy − g·t)²).
Check that your calculator is set to degrees before you take sin or cos of an angle given in degrees.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
A ball is thrown horizontally at 12 m/s from the top of a 45 m cliff. Ignore air resistance and use g = 9.81 m/s². How far from the foot of the cliff does it land?
1
Draw the situation
The launch is horizontal, so vx = 12 m/s and vy = 0. The ball must fall h = 45 m.
2
Find the time of flight
The vertical motion is a fall from rest through 45 m:
t = √(2h / g) = √(2 · 45 / 9.81) = √9.174 = 3.03 s
3
Use the time sideways
During those 3.03 s the ball keeps moving sideways at 12 m/s:
x = vx · t = 12 · 3.029 = 36.3 m
4
Check the answer
Put the time back into the vertical motion: ½ · 9.81 · 3.029² = 45.0 m. That is exactly the height of the cliff, so the time is right.
A distance of about 36 m for a 12 m/s throw over 3 s is also a sensible size.
5
Going further: the impact speed
Just before landing, vy = g · t = 9.81 · 3.029 = 29.7 m/s downwards, while vx is still 12 m/s.
v = √(12² + 29.7²) = 32.0 m/s
Answer
The ball lands x ≈ 36.3 m from the foot of the cliff, after 3.03 s.
6Common mistakes
Using the full speed v0 where a component belongs. The time of flight comes from vy only, the range from vx only.
Mixing up sin and cos. The horizontal component goes with cos θ when θ is measured from the horizontal.
A calculator in radians while the angle is in degrees.
Using R = v0² · sin 2θ / g when the ball lands lower or higher than it started. Then you must solve the vertical equation instead.
Giving the horizontal motion an acceleration. Sideways, the speed is constant.
Self-check A ball is kicked from level ground at 20 m/s and 30° above the horizontal. How long is it in the air? (g = 9.81 m/s²)
1.02 s
2.04 s
3.53 s
4.08 s
Answer: 2.04 s. vy = 20 · sin 30° = 10 m/s, so t = 2 · 10 / 9.81 = 2.04 s. Using the full 20 m/s gives 4.08 s, and 1.02 s is only the time to the highest point.
7How exam questions look
Easy: a horizontal launch from a cliff or wall. Find the time of flight, the landing distance, the launch speed from a given landing distance, or the impact speed. Some easy questions ask for the time to the highest point of a launch from the ground.
Medium: a launch from level ground at an angle. Find the time of flight, the maximum height or the range. Variations ask for the launch speed or angle that gives a certain range, or the height at which the ball passes a wall.
Hard: a launch at an angle from the top of a cliff, or a landing on a roof. Setting y = 0 (or the roof height) gives a quadratic equation in t: take the positive root.
In the hard case, the landing angle below the horizontal follows from the final components: tan α = |vy − g·t| / vx.
A ball is launched horizontally from the top of a 55 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
v0 = 10 m/sθ = 0°h = 55 mg = 9.81 m/s²
Find x(m)
Show hints
Treat horizontal and vertical motion separately. They share only the time t.
The ball lands when its height y(t) reaches 0. Write y(t) = h + vyt − ½gt².
vy = 0 m/s, and the flight lasts 3.349 s.
Show the answer and the step-by-step solution
Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.
Split the launch velocity
Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.
vx = v0·cos θ = 10·cos 0° = 10 m/s
vy = v0·sin θ = 10·sin 0° = 0 m/s
Time in the air
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
t = √(2h/g) = √(2·55/9.81) = 3.349 s
Horizontal distance
x = vx·t = 10·3.349 = 33.49 m
Answer
x = 33.49 m
Example 2Medium
A ball is launched at 60° above horizontal from level ground. Ignore air resistance. Find the time of flight t until the ball hits the ground.
v0 = 26 m/sθ = 60°g = 9.81 m/s²
Find t(s)
Show hints
Treat horizontal and vertical motion separately. They share only the time t.
By symmetry, time up equals time down.
vy = 22.52 m/s.
Show the answer and the step-by-step solution
Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.
Split the launch velocity
Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.
vx = v0·cos θ = 26·cos 60° = 13 m/s
vy = v0·sin θ = 26·sin 60° = 22.52 m/s
Time in the air
Starting and landing at the same height: t = 2vy/g.
t = 2·22.52/9.81 = 4.591 s
Answer
t = 4.591 s
Example 3Hard
A ball is launched at 25° above horizontal from the top of a 55 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
v0 = 33 m/sθ = 25°h = 55 mg = 9.81 m/s²
Find t(s)
Show hints
Treat horizontal and vertical motion separately. They share only the time t.
The ball lands when its height y(t) reaches 0. Write y(t) = h + vyt − ½gt².
vy = 13.95 m/s.
Show the answer and the step-by-step solution
Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.
Split the launch velocity
Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.
vx = v0·cos θ = 33·cos 25° = 29.91 m/s
vy = v0·sin θ = 33·sin 25° = 13.95 m/s
Time in the air
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
0 = 55 + 13.95·t − 4.905·t²
t = (vy + √(vy² + 2gh))/g = (13.95 + √(194.5 + 1079))/9.81 = 5.06 s
Answer
t = 5.06 s
Questions and answers
What is Projectile motion?
A projectile is an object that flies through the air with only gravity acting on it. A kicked ball, a stone thrown from a bridge and water from a hose are all projectiles.
In this topic you ignore air resistance. Then the path is always a smooth curve called a parabola.
How do I solve Projectile motion problems step by step?
Split the launch velocity into components: vx = v₀·cos θ and vy = v₀·sin θ.
The horizontal motion has constant speed; the vertical motion has acceleration −g.
Find the time of flight from the vertical motion.
Use that time in the horizontal motion to find the range.
What are common mistakes in Projectile motion?
Using v₀ instead of a component.
Calculator in radians when the angle is in degrees.
Using the range formula v₀²·sin 2θ/g when the launch and landing heights differ.
What do the Projectile motion exercises cover?
Easy: horizontal launch
Medium: launch from ground
Hard: launch from height
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.