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.

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

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 by time:

Horizontal: $u_x = u\cos\theta$ , constant, so $x = (u\cos\theta)\,t$ .
Vertical: $u_y = u\sin\theta$ initially, then gravity decelerates it on the way up

and accelerates it down:

v_y = u\sin\theta - gt

and

y = (u\sin\theta)t - \tfrac12 gt^2

(taking up as positive).Three landmarks fall straight out of the vertical motion:

Greatest height: the vertical velocity is momentarily zero ( $v_y = 0$ ). Solve

0 = u\sin\theta - gt

for the time, or use

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

Time of flight (back to launch level): the vertical displacement returns to zero,

giving

t = \dfrac{2u\sin\theta}{g}

Range: the horizontal distance in that time, $x = (u\cos\theta)\,t$ .

The path is again a parabola, now symmetric about its highest point.