The idea

Projection at an angle

Resolving the launch velocity into u cosθ (horizontal, constant) and u sinθ (vertical, under gravity), then using suvat in each direction to find the time of flight, greatest height and range.

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Mechanics · Projectiles

Projection at an angle

Resolving the launch velocity into u cosθ (horizontal, constant) and u sinθ (vertical, under gravity), then using suvat in each direction to find the time of flight, greatest height and range.

Why it works

Split the launch velocity

When a body is launched at an angle

\theta

to the horizontal with speed

u

, split that launch velocity into two components — and then it behaves exactly like the horizontal case, with the two directions independent and linked only by time:

u_x = u\cos\theta, \qquad u_y = u\sin\theta

Greatest height: the vertical pause

At the top of the arc the ball is still moving — but only sideways. The vertical velocity is momentarily zero, and that single fact unlocks the height: solve

0 = u\sin\theta - gt

for the time to the top, or go straight to the height with

v_y^2 = (u\sin\theta)^2 - 2gH

and

v_y = 0

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