Projectile Motion Calculator

This projectile motion calculator finds the flight of anything launched through the air: a ball, a stone, water from a hose. Enter the launch speed, the launch angle and the starting height, and it returns the horizontal and vertical velocity components, the time of flight, the range (horizontal distance), the maximum height and when it’s reached, and the speed and angle at impact.

It assumes no air resistance and constant gravity, the standard model in physics and calculus courses. It’s free and needs no sign-up.

How to use the projectile motion calculator

1. Enter the launch speed \( v_0 \) in m/s.
2. Enter the launch angle \( \theta \) in degrees above the horizontal. Use 0 for a horizontal launch.
3. Enter the launch height \( h_0 \) in metres above the landing level (0 if it starts on the ground).
4. Leave g at 9.81 m/s² for Earth, or change it (1.62 for the Moon). Press Launch.

Formulas

Split the launch velocity into components:

$$v_x = v_0\cos\theta, \qquad v_y = v_0\sin\theta$$

Horizontally nothing pushes the projectile, so \( x(t) = v_x t \). Vertically gravity pulls it down:

$$y(t) = h_0 + v_y t - \tfrac12gt^2$$

From these, for a launch from ground level (\( h_0 = 0 \)):

$$T = \frac{2v_0\sin\theta}{g}, \qquad R = \frac{v_0^2\sin 2\theta}{g}, \qquad H = \frac{v_0^2\sin^2\theta}{2g}$$

For \( h_0 > 0 \) the calculator solves \( y(T) = 0 \) with the quadratic formula instead.

Worked example: a 45° launch

A ball is kicked at 20 m/s at \( 45^\circ \) from the ground, as loaded above.

Components. \( v_x = v_y = 20\cos 45^\circ \approx 14.142 \) m/s.

Time of flight. \( T = \frac{2(14.142)}{9.81} \approx 2.883 \) s.

Range. \( R = \frac{20^2\sin 90^\circ}{9.81} = \frac{400}{9.81} \approx 40.77 \) m.

Maximum height. \( H = \frac{14.142^2}{2(9.81)} \approx 10.19 \) m, reached at \( t \approx 1.44 \) s, halfway through the flight.

Worked example: horizontal launch from a height

A ball rolls off a 20 m cliff at 15 m/s. With \( \theta = 0 \), \( v_y = 0 \), so the fall time comes from \( 20 = \frac12(9.81)t^2 \):

$$T = \sqrt{\frac{2 \cdot 20}{9.81}} \approx 2.019 \text{ s}, \qquad R = 15 \cdot 2.019 \approx 30.29 \text{ m}$$

At impact the vertical speed is \( 9.81 \cdot 2.019 \approx 19.81 \) m/s, so the impact speed is \( \sqrt{15^2 + 19.81^2} \approx 24.85 \) m/s, about \( 52.9^\circ \) below horizontal.

Worked example: launch from above the ground

Thrown at 20 m/s and \( 60^\circ \) from a hand 1.5 m up: \( v_x = 10 \) m/s and \( v_y \approx 17.32 \) m/s. Solving \( 1.5 + 17.32t - 4.905t^2 = 0 \) gives \( T \approx 3.616 \) s, so \( R \approx 36.16 \) m. The maximum height is \( 1.5 + \frac{17.32^2}{2(9.81)} \approx 16.79 \) m.

Tips and common mistakes

• Degrees vs radians. The angle here is in degrees. A hand calculator in radian mode will give nonsense for \( \cos 45 \).
• Using the range formula from a height. \( R = \frac{v_0^2\sin 2\theta}{g} \) only holds when launch and landing heights are equal.
• 45° is best only on level ground. From a height, the angle that maximizes range is a little less than \( 45^\circ \).
• Complementary angles. On level ground, \( 30^\circ \) and \( 60^\circ \) give the same range because \( \sin 60^\circ = \sin 120^\circ \).

The calculus behind it

Gravity gives constant acceleration \( a_y = -g \). Integrating once gives \( v_y(t) = v_0\sin\theta - gt \), and setting it to zero gives the time of maximum height. Integrating again gives \( y(t) \). Maximizing \( R(\theta) = \frac{v_0^2\sin 2\theta}{g} \) by setting \( R'(\theta) = 0 \) gives \( \theta = 45^\circ \), a classic optimization problem. The projectile motion guide covers all of this in detail.

Related guides

• Projectile Motion: Equations and Examples
• Kinematic Equations
• Optimization Problems
• Trigonometric Functions

Further reading

• 4.3 Projectile Motion (OpenStax University Physics Volume 1) — derivations of time of flight, range and trajectory with examples.
• Projectile motion (Wikipedia) — equations, the parabolic path and the effect of air resistance.

FAQ

What angle gives the maximum range?

\( 45^\circ \) when the projectile lands at the same height it was launched from, ignoring air resistance.

How do I find the time of flight?

On level ground, \( T = \frac{2v_0\sin\theta}{g} \). From a height, solve \( h_0 + v_0\sin\theta\,t - \frac12gt^2 = 0 \) for the positive t.

Does mass affect projectile motion?

Not without air resistance. A heavy and a light ball launched the same way follow the same path.

Why is the path a parabola?

\( x \) grows linearly with time and \( y \) is quadratic in time, so eliminating t gives y as a quadratic function of x.