Projectile Motion: Numerical Problems Set (5)

Question

A ball is projected at $u = 50 \text{ m/s}$ at $\theta = 53°$ above the horizontal from ground level. Find (a) time of flight, (b) maximum height, (c) horizontal range, (d) speed and direction at $t = 4 \text{ s}$. Take $g = 10 \text{ m/s}^2$, $\sin 53° = 0.8$, $\cos 53° = 0.6$.

Solution — Step by Step

Final answers: $T = \mathbf{8 \text{ s}}$, $H = \mathbf{80 \text{ m}}$, $R = \mathbf{240 \text{ m}}$, at $t = 4 \text{ s}$: $v = \mathbf{30 \text{ m/s}}$ horizontal.

Why This Works

Projectile motion is two independent 1D motions glued together. The horizontal direction is uniform ($a_x = 0$); the vertical is uniformly accelerated ($a_y = -g$). Time is the shared variable.

The peak occurs at $t = u_y/g = T/2$ — exactly half the time of flight. Our $t = 4 \text{ s}$ being half of $8 \text{ s}$ is why $v_y = 0$ there. Knowing this saves arithmetic.

Alternative Method

Using the trajectory equation $y = x\tan\theta - gx^2/(2u^2\cos^2\theta)$ and the range formula directly: $R = u^2\sin 2\theta/g$, $H = u^2\sin^2\theta/(2g)$. These are quicker for MCQs but obscure the kinematic logic.

Common Mistake