Projectiles launched at an angle

5 minAS Level2.1.72.1.9

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can solve problems on projectiles launched at an angle: resolve the launch velocity, then find the time of flight, the maximum height, the horizontal range and the velocity at any point.

Key to the symbols

The main symbols used in this lesson:

Symbol	Meaning
$\theta$	the angle above the horizontal
$u_x$	horizontal velocity (it stays the same)

Same trick, one extra step

A goal kick, a basketball shot, a stone lobbed off a cliff: the launch velocity now points up at an angle $\theta$ . So the first move is always the same. Resolve it:

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

After that, it's last lesson's two-column method, with one change: the vertical motion now starts with $u_y$ upwards instead of zero.

The two-column recipe

Taking upwards as positive:

	horizontal	vertical
initial velocity	$u\cos\theta$	$u\sin\theta$
acceleration	$0$	$-9.81\ \text{m s}^{-2}$
velocity at time $t$	$u\cos\theta$ , always	$u\sin\theta - 9.81t$
displacement	$x = (u\cos\theta)\,t$	$y = (u\sin\theta)\,t - \tfrac{1}{2}(9.81)t^2$

Three checkpoints turn up in almost every question:

At the top: the vertical velocity is zero. The velocity is not zero: it's $u\cos\theta$ , horizontal.
Landing at launch height: $y = 0$ , and the flight is symmetrical, so time up = time down.
Velocity at any point: combine $v_x$ and $v_y$ with Pythagoras; the direction is $\tan^{-1}\left(\dfrac{v_y}{v_x}\right)$ .

Quick check

Multiple choiceQuestion 1 · Recall

Answer the 2 questions above to continue.