Physics · Units 3 & 4

Projectile Motion

Understand projectile motion the easy way, with plain English intuition, an interactive simulation, the independence of horizontal and vertical motion, worked examples and an auto marked practice test. VCE Physics Units 3 and 4.

Learn

Throw a ball and it carves a smooth arch through the air, every single time. That curved path is a projectile, and the secret to predicting exactly where it lands is almost absurdly simple. You split the motion into two completely separate problems, one going across and one going up and down, and solve each on its own. Gravity only ever pulls straight down, so the across and the up and down never interfere with each other.

Two motions, side by side

Here is the whole trick. A projectile is doing two things at once, and they do not talk to each other:

  • Horizontally there is no force (we ignore air resistance), so the sideways velocity is constant. The ball drifts across at a steady speed.
  • Vertically gravity pulls down at 9.89.8 m/s2^2, exactly as if you had simply dropped the ball. It slows on the way up, stops for an instant, then speeds up on the way down.

The path you see is just these two simple motions happening together.

uθvxvymaximum heightrange

The blue arrows are the horizontal velocity, and notice they stay the same length all the way along. The red arrow is the vertical velocity, which grows as the ball falls. The path bends only because the vertical motion keeps changing while the horizontal one does not.

The two motions share exactly one thing: time

This is the idea that unlocks every projectile question. The horizontal problem and the vertical problem are separate, but they happen over the same stretch of time. So you use the vertical motion to find the time in the air, then feed that time into the horizontal motion to find how far it travelled.

See it for yourself

Drag the cannon, change the launch speed and angle, and watch the horizontal and vertical motion play out. Try setting the angle to 45∘45^\circ to get the longest range, and notice the horizontal speed never changes while the ball is in the air.

Interactive simulation, Projectile Motion Source: PhET Interactive Simulations, University of Colorado Boulder (CC BY 4.0)

How to actually solve one

The method is a short, repeatable recipe.

  1. Split the launch velocity into ux=ucos⁡θu_x = u\cos\theta and uy=usin⁡θu_y = u\sin\theta.
  2. Use the vertical motion to find the time. At the top, vy=0v_y = 0. To land back at launch height, the flight time is T=2uygT = \dfrac{2u_y}{g}.
  3. Use the horizontal motion with that time: x=ux tx = u_x\,t.
  4. For maximum height, use h=uy22gh = \dfrac{u_y^{2}}{2g}.

Watch the signs. If you call up positive, then gravity is −9.8-9.8 m/s2^2, and a stone thrown horizontally simply starts with uy=0u_y = 0.

See the recipe in action in the Worked Examples tab, then test yourself in Try It.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

How do you split a launch velocity uu at angle θ\theta into components?
What is the only quantity the horizontal and vertical motions share?
Why does the horizontal speed stay constant during the flight?
What is the acceleration at the very top of a projectile’s path?
Write the formula for the maximum height reached from vertical launch speed uyu_y.
Recall · Newton's Laws and Forces
State Newton’s second law.
Recall · Uniform Circular Motion
In what direction does the acceleration of an object in uniform circular motion point?

Worked examples

Worked Example 1A ball kicked at an angle

A ball is kicked from ground level at 2020 m/s, 30∘30^\circ above the horizontal. Taking g=9.8g = 9.8 m/s2^2 and ignoring air resistance, find the time of flight, the maximum height and the horizontal range.

  1. 1

    Split the launch velocity into a horizontal part and a vertical part. These run independently from now on.

    ux=20cos⁡30∘≈17.3 m/s,uy=20sin⁡30∘=10 m/su_x = 20\cos 30^\circ \approx 17.3 \text{ m/s}, \qquad u_y = 20\sin 30^\circ = 10 \text{ m/s}
  2. 2

    The vertical motion sets the clock. At the top the vertical velocity is zero, so the time up is uy/gu_y / g, and the flight takes twice that.

    tup=uyg=109.8=1.02 s,T=2 tup=2.04 st_{\text{up}} = \frac{u_y}{g} = \frac{10}{9.8} = 1.02 \text{ s}, \qquad T = 2\,t_{\text{up}} = 2.04 \text{ s}
  3. 3

    Maximum height uses the vertical part only.

    h=uy22g=1022×9.8=5.10 mh = \frac{u_y^{2}}{2g} = \frac{10^{2}}{2 \times 9.8} = 5.10 \text{ m}
  4. 4

    The horizontal velocity never changes, so the range is just horizontal speed times the total time.

    R=ux T=17.3×2.04=35.3 mR = u_x\,T = 17.3 \times 2.04 = 35.3 \text{ m}
Answer
T=2.04 s,h=5.10 m,R=35.3 mT = 2.04 \text{ s}, \quad h = 5.10 \text{ m}, \quad R = 35.3 \text{ m}
Worked Example 2Thrown horizontally off a cliff

A stone is thrown horizontally at 1515 m/s from the top of a 2020 m high cliff. Taking g=9.8g = 9.8 m/s2^2, find the time to reach the ground and the distance from the base of the cliff where it lands.

  1. 1

    Thrown horizontally means the starting vertical velocity is zero. Vertically the stone simply falls.

    uy=0,y=12gt2u_y = 0, \qquad y = \tfrac{1}{2} g t^{2}
  2. 2

    Find the time to fall 2020 m.

    20=12(9.8) t2  ⟹  t=409.8=2.02 s20 = \tfrac{1}{2}(9.8)\,t^{2} \implies t = \sqrt{\tfrac{40}{9.8}} = 2.02 \text{ s}
  3. 3

    The horizontal velocity stays 1515 m/s the whole way, so distance is speed times time.

    x=ux t=15×2.02=30.3 mx = u_x\,t = 15 \times 2.02 = 30.3 \text{ m}
Answer
t=2.02 s,x=30.3 mt = 2.02 \text{ s}, \quad x = 30.3 \text{ m}

Practice questions

Practice test

Try it yourself

6 questions, 11 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.A ball follows a curved projectile path. At the highest point of its flight, ignoring air resistance, its acceleration is:

1mark
Need a hint?
Ask what force acts on the ball in the air. Acceleration follows the force, not the velocity.

Q2.One ball is launched horizontally off a table at the exact instant an identical ball is dropped from the same height. Ignoring air resistance, which ball lands first?

1mark
Need a hint?
The horizontal and vertical motions are independent. Only the vertical motion decides the fall time.

Q3.A projectile is launched at 2525 m/s, 53∘53^\circ above the horizontal. Using sin⁡53∘≈0.80\sin 53^\circ \approx 0.80, the vertical component of its launch velocity is closest to:

1mark
Need a hint?
The vertical component is usin⁡θu\sin\theta.

Q4.A ball leaves the ground with a vertical velocity component of 1414 m/s. Using g=9.8g = 9.8 m/s2^2, the maximum height it reaches is closest to:

1mark
Need a hint?
Use h=uy22gh = \dfrac{u_y^{2}}{2g}.

Q5.A stone is thrown horizontally at 8.08.0 m/s from the top of a 5.05.0 m high wall. Taking g=9.8g = 9.8 m/s2^2, find the time it takes to reach the ground and the horizontal distance it travels. Show your working.

3marks

Work this on paper. The worked solution appears once you submit.

Q6.A cat gently taps a small ball off a ledge so it drops vertically to the floor 1.801.80 m below. Air resistance is ignored (g=9.8g = 9.8 m/s2^2). (a) Show that the ball takes 0.6060.606 s to reach the floor. (b) Calculate the speed of the ball just before it hits the floor. (c) On a second try the ball instead leaves the ledge horizontally at 0.750.75 m/s. State whether the time to reach the floor is less than, the same as, or greater than in part (a), and justify your answer with no calculation.

4marks

Work this on paper. The worked solution appears once you submit.

VCAA 2025 Physics Exam, Section B Q1

Frequently asked questions

Why does the horizontal speed stay the same?
Ignoring air resistance, the only force on a projectile is gravity, which acts straight down. There is no horizontal force, so by Newton's first law the horizontal velocity never changes for the whole flight.
Is the acceleration zero at the top of the path?
No. The vertical velocity is zero for an instant at the very top, but the acceleration stays 9.8 m/s squared downwards the entire time, because gravity never switches off.
Do heavier projectiles fall faster?
No. Ignoring air resistance, every projectile accelerates downward at the same 9.8 m/s squared regardless of mass, so a heavy ball and a light ball launched the same way trace the same path.