Teaching

Projectile motion

A kicked ball, a jet of water, a stone thrown from a bridge: they all follow the same law. Set up the throw, then compare trajectories.

Controls

Launch angle (°)45Initial speed (m/s)10Launch height (m)0Gravity

Current throw

Results

Flight time: 1.44 s
Range: 10.2 m
Maximum height: 2.55 m
Impact speed: 10.0 m/s

The equations

Axes: origin on the ground, directly below the launch point; x-axis horizontal, y-axis pointing up.

Equations of motion

x(t) = v₀ cos α · t

y(t) = −½ g t² + v₀ sin α · t + h₀

x(t) = 7.07 t

y(t) = −4.90 t² + 7.07 t

Trajectory (a parabola)

y = −g x² / (2 v₀² cos² α) + x tan α + h₀

y = −0.0980 x² + 1.00 x

Maximum height

H = h₀ + v₀² sin² α / (2g)

H = 2.55 m

Flight time

t_v = [v₀ sin α + √(v₀² sin² α + 2 g h₀)] / g

t_v = 1.44 s

Range

R = v₀ cos α · t_v

with no launch height, R = v₀² sin 2α / g

R = 10.2 m

Impact speed

v = √(v₀² + 2 g h₀)

v = 10.0 m/s

Assumptions

The projectile is treated as a point mass, and the Earth’s frame of reference as inertial. Air resistance is neglected: only its weight acts on it, so its acceleration is g, pointing straight down.

Try this

With no launch height, throw at 30° and then at 60° at the same speed. What do you notice?
Show the explanation
the range is v₀² sin 2α / g, and sin 60° = sin 120°. Two complementary angles give the same range.
Still with no launch height, find the angle that gives the longest range.
Show the explanation
45°, because sin 2α is at most 1, when 2α = 90°.
Make the same throw on Earth and then on the Moon, with no launch height. By how much is the range multiplied?
Show the explanation
by 9.8 / 1.6 ≈ 6.1. With no launch height, the range is inversely proportional to g.