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Projectile motion calculator

A projectile motion calculator finds how far a thrown or launched object travels, how high it goes and how long it stays in the air. Split the launch velocity into a steady horizontal part and a vertical part that gravity slows, and the rest follows: on level ground, range = v² sin 2θ ÷ g. Our Projectile Motion Simulator launches it for you and reads off the range, maximum height, flight time and impact speed.

It opens on a launch at 20 m/s and 45° from the ground, which lands 40.8 m away. Turn on air resistance to see how a real ball falls short of the textbook answer.

Try it

Projectile Motion Simulator

Launch projectiles and read off range, height and flight time

Go to its page

How to use the projectile motion simulator

  1. Under Launch, set the launch speed (1 to 300 m/s), the angle above the horizontal (0° to 90°) and the launch height (0 to 1,000 m). You can also drag the launch arrow on the stage.
  2. Pick the gravity: Earth, the Moon, Mars or Jupiter, or type your own value.
  3. Leave Air resistance off for textbook answers. Turn it on to choose a football, tennis ball, golf ball or cannonball, or set the mass, diameter and drag coefficient yourself.
  4. Press play under Run. The stage marks the range and the maximum height, and the readouts give the flight time, the impact speed and the impact angle.
  5. Pause the flight, or step it forward, and read the time, the distance across, the height and both parts of the velocity at that moment. You can slow it down too.
  6. Under Compare launches, press Keep this launch, change a setting and launch again. Up to five launches stay on the stage with a table of their range, height and time.

Readings are given to 3 significant figures. Practice and help sets range, maximum height, flight time and cliff questions with worked solutions. The simulator shows results, not working, so the examples below show each step.

Projectile motion formulas

Ignore air resistance and nothing pushes the projectile sideways, so its sideways speed never changes, while gravity takes g off its upward speed every second. For a launch at speed v and angle θ:

v_x = v\cos\theta \qquad v_y = v\sin\theta
To findLevel groundAny launch
Time to the topt = v sin θ ÷ gt = v sin θ ÷ g
Maximum heightH = (v sin θ)² ÷ 2gLaunch height + (v sin θ)² ÷ 2g
Flight timeT = 2v sin θ ÷ gSolve h + v sin θ t − ½gt² = 0
RangeR = v² sin 2θ ÷ gR = v cos θ × T

The level-ground formulas only work when the projectile lands at the height it was launched from. For anything else, work out the flight time from the vertical motion first, then multiply by the horizontal velocity. This uses the SUVAT equations in each direction separately.

25 m/s21.7 m/s12.5 m/s30°
The launch velocity splits into a horizontal part and a vertical part

Worked examples you can check in the simulator

Example 1: range on level ground

A ball is kicked at 20 m/s, 45° above the horizontal, from the ground. Take g = 9.81 m/s²:

R = \frac{20^2 \sin 90°}{9.81} = 40.8\ \text{m} \qquad H = \frac{(20\sin 45°)^2}{2 \times 9.81} = 10.2\ \text{m}

The flight time is 2 × 20 sin 45° ÷ 9.81 = 2.88 s. These are the simulator's opening values, and it gives range 40.8 m, maximum height 10.2 m and flight time 2.88 s.

y = x − 0.0245x²
510152025303540-224681012Top, 10.2 mLands, 40.8 m
Drag to pan, scroll to zoom. x from 0 to 42, y from -3.55 to 12.5.

Here x is the distance across in metres and y is the height in metres. The path is a parabola, from y = x tan θ − gx² ÷ (2v² cos² θ). The two axes are drawn to different scales, so the path looks steeper than it is.

Example 2: a horizontal launch off a cliff

A stone is thrown horizontally at 15 m/s off a 45 m cliff. With θ = 0 the vertical velocity starts at zero, so the time comes from the drop alone:

45 = \frac{1}{2} \times 9.81 \times t^2 \Rightarrow t = 3.03\ \text{s}
\text{distance} = 15 \times 3.03 = 45.4\ \text{m}

It lands with a vertical velocity of 9.81 × 3.03 = 29.7 m/s, so its impact speed is √(15² + 29.7²) = 33.3 m/s. Set speed 15, angle 0 and height 45: the simulator gives 45.4 m, 3.03 s, an impact speed of 33.3 m/s and an impact angle of 63.2° below the horizontal.

Try one in the course, then press Mark:

Find the flight time or the landing distance of a ball fired sideways off a cliff.

Press Mark to have your answer marked in the Laws of Motion course.

Example 3: an angled launch from a height

A ball is thrown at 25 m/s, 30° above the horizontal, from 10 m up. First the parts of the velocity: v_x = 25 cos 30° = 21.65 m/s and v_y = 25 sin 30° = 12.5 m/s. Taking up as positive, it lands when its height is zero:

10 + 12.5t - 4.905t^2 = 0 \Rightarrow t = \frac{12.5 + \sqrt{12.5^2 + 2 \times 9.81 \times 10}}{9.81} = 3.19\ \text{s}

The range is 21.65 × 3.19 = 69.0 m, and the maximum height is 10 + 12.5² ÷ 19.62 = 18.0 m. The simulator gives the same three numbers. Using R = v² sin 2θ ÷ g here would give 55.2 m, which is wrong because the ball lands lower than it started.

Example 4: two angles, one range

Launch at 20 m/s and 30°, keep it, then launch at 60°. Both land 35.3 m away, because sin 60° = sin 120°. The 60° launch climbs to 15.3 m and takes 3.53 s; the 30° launch reaches only 5.10 m in 2.04 s. On level ground with no air, 45° gives the greatest range, and any pair of angles adding up to 90° give equal ranges.

Air resistance and terminal velocity

Turn on air resistance and the drag force, ½ρCdAv², slows the ball in both directions. A football kicked at 25 m/s and 45° lands 63.7 m away with no air, but only 39.6 m away with air, using the simulator's football: 0.43 kg, 22 cm across, drag coefficient 0.25. The path is no longer a parabola: it comes down more steeply than it went up. With air on, the best angle is lower than 45°: at 40° this football goes 39.8 m.

A falling object stops speeding up when drag balances its weight. That speed is its terminal velocity:

v_t = \sqrt{\frac{2mg}{\rho C_d A}}

For the football, A = π × 0.11² = 0.0380 m², and v_t = √(2 × 0.43 × 9.81 ÷ (1.225 × 0.25 × 0.0380)) = 26.9 m/s. The simulator has no terminal velocity readout, but you can see it: drop the football from 1,000 m at 1 m/s and 0° with air on, and it hits the ground at 26.9 m/s. The tennis ball reaches only 21.9 m/s.

Projectile motion on other planets

Choose Moon and launch at 20 m/s and 45° again. With g = 1.62 m/s² and no air, the ball goes 247 m, six times as far as on Earth, and stays up for 17.5 s. Range is inversely proportional to g, so a weaker pull means a longer flight.

What the simulator does not do

  • It works forwards only. It finds the range from the launch, not the launch speed or angle needed to hit a target. Rearrange by hand, then check your answer by launching it.
  • No launches below the horizontal. The angle runs from 0° to 90°, so a ball thrown downwards needs working by hand.
  • No working shown. It gives results to 3 significant figures, so set out the steps yourself, as in the examples.

Common mistakes

  • Using the level-ground range formula for every launch. R = v² sin 2θ ÷ g only works if the projectile lands at its launch height.
  • Putting gravity into the horizontal motion. With no air, nothing acts sideways, so v_x stays the same all flight.
  • Mixing up sin and cos. With θ measured from the horizontal, v_x = v cos θ and v_y = v sin θ.
  • Saying the ball stops at its highest point. Only its upward motion stops there. It is still moving sideways at v cos θ.
  • Calculator in radians. Check your calculator is in degree mode before you take sin 30°.

Practise projectiles in a short timed paper

Projectile motion comes up in AP Physics 1 and AP Physics C: Mechanics practice, and in A-level Maths: Mechanics and A-level Physics: Year 1 in the UK. Each daily paper is a short timed paper of 8 questions worth 25 to 35 marks, and the papers mix topics. For one-direction motion, use the SUVAT Calculator.

Common questions

How do you calculate the range of a projectile?
On level ground with no air resistance, R = v² sin 2θ ÷ g. Otherwise, find the flight time from the vertical motion and multiply it by the horizontal velocity, v cos θ.
What angle gives the maximum range?
45° on level ground with no air resistance. With air resistance the best angle is lower, and launching from a height also lowers it.
How do you find the maximum height of a projectile?
Use H = (v sin θ)² ÷ 2g, and add the launch height if it starts above the ground.
How do you find the time of flight?
On level ground, T = 2v sin θ ÷ g. From a height, solve h + v sin θ t − ½gt² = 0 for t, taking the positive answer.
Does mass affect projectile motion?
Not without air resistance: every mass falls with the same acceleration. With air resistance, a heavier ball of the same size and shape slows less and goes further.
How do you calculate terminal velocity?
v_t = √(2mg ÷ ρC_dA), where ρ is the air density, C_d the drag coefficient and A the cross-sectional area. The simulator uses this drag model but does not display v_t.

Revision cards

Say the answer first, then flip the card to check it.

Where to go next

  • Solve it

    Projectile Motion Simulator

    Launch projectiles and read off range, height and flight time

    Open the tool
  • Learn it

    Laws of Motion

    Newton's laws, momentum, and the forces that shape our world

    Start the course
  • Test it

    AP Physics 1

    A new exam-style paper every day, with instant marking and a mark scheme

    Start today's paper
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