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Projectile motion

When something is thrown at an angle, the motion splits into a horizontal part with constant velocity and a vertical part with free fall. The two directions do not affect each other, so you solve them separately and connect them only through the time tt.

x=v0cos⁡θ⋅tx = v_0\cos\theta\cdot thorizontal position
y=v0sin⁡θ⋅t−12gt2y = v_0\sin\theta\cdot t - \tfrac12 gt^2vertical position (upward positive)

Symbols

v0v_0initial speedm/s
θ\thetalaunch angle°
x, yx,\ yhorizontal/vertical positionm

Example

A ball is thrown at v0=15v_0 = 15 m/s at θ=30∘\theta = 30^\circ. Horizontal component: v0cos⁡30∘≈13v_0\cos 30^\circ \approx 13 m/s. After t=1.0t = 1.0 s: x≈13x \approx 13 m.

Always split into an x- and a y-component first – time is the only thing that connects them.
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Part of Foundations of Physics: Motion.