Projectile Motion: Range, Maximum Height, Time of Flight

From the initial speed and the launch angle (degrees), finds the horizontal range, the maximum height and the time of flight. Speed is in m/s, distances in metres, and time in seconds. Air resistance is ignored.

Explanation

Launch something at an angle and this works out how far it travels, how high it climbs and how long it stays up, from the moment of launch until it comes back down to the height it started from.

R = \dfrac{v_0^2 \sin 2\theta}{g}

H = \dfrac{(v_0 \sin\theta)^2}{2g}

T = \dfrac{2 v_0 \sin\theta}{g}

$R$ is the range and $H$ the maximum height, both in metres (m)
$T$ is the time of flight, in seconds (s)
$v_0$ is the launch speed, in m/s
$\theta$ is the launch angle, entered here in degrees
$g$ is the acceleration of gravity, about 9.8 m/s²

Nothing pushes the projectile sideways, so the horizontal motion runs at constant speed while the vertical motion is pure free fall. Treat the two separately and the formulas above drop out.

Example

The defaults are a launch speed of 20 m/s, an angle of 45 degrees and gravity of 9.8 m/s².

R = \dfrac{20^2 \times \sin 90^\circ}{9.8} = \dfrac{400}{9.8} \approx 40.8\,\mathrm{m}

The range is about 40.8 m, the maximum height about 10.2 m, and the flight lasts about 2.89 s.

Notes

Air resistance is ignored, so a real baseball or golf ball, slowed by drag and steered by spin, will not match these numbers.

The range is greatest at 45 degrees. For a given speed, 30 degrees and 60 degrees give exactly the same range, because $\sin 2\theta$ works out the same for both.

The formulas assume the landing height equals the launch height. Throwing from the top of a cliff is a different problem.

Angles go in degrees here, not radians.