Projectile Motion: Real-World Scenarios (4)

Question

A cricket ball is thrown from ground level at $u = 20$ m/s at an angle $\theta = 45°$ above the horizontal. Find (a) the time of flight, (b) the maximum height reached, (c) the horizontal range, and (d) the velocity (magnitude and direction) when the ball is at half its maximum height on the way up. Take $g = 10$ m/s$^2$.

Solution — Step by Step

Final answers: $T \approx 2.83$ s; $H = 10$ m; $R = 40$ m; $v \approx 17.32$ m/s at $35.3°$ above horizontal.

Why This Works

Projectile motion splits cleanly into independent horizontal and vertical motions. Horizontal: constant velocity. Vertical: uniform acceleration $-g$. Once you internalise this, every projectile question becomes two 1D problems running in parallel.

The ”$45°$ gives maximum range” result drops out of $R = u^2 \sin 2\theta / g$ — $\sin 2\theta$ peaks at $2\theta = 90°$, i.e. $\theta = 45°$. Useful for elimination on MCQs.

Alternative Method

Use energy conservation for parts (b) and (d). At max height, all vertical KE has converted to PE: $\frac{1}{2}m u_y^2 = m g H \Rightarrow H = u_y^2 / (2g) = 10$ m. At half height, vertical KE = $\frac{1}{2}m u_y^2 - m g (H/2) = \frac{1}{2}m u_y^2 / 2$. So $v_y = u_y/\sqrt{2} \approx 10$ m/s. Same answer.

Common Mistake

Treating $g$ as $9.8$ in some steps and $10$ in others. Pick one at the start and stick with it. JEE/NEET usually allow either; the marking accepts both consistent computations.