1. Write the projectile motion equations

A ball is projected with speed $v_0$ at angle $\theta$.

Its horizontal and vertical positions at time $t$ are:

$x = v_0 \cos\theta\, t$ $y = v_0 \sin\theta\, t - \frac{1}{2}gt^2$

2. Condition for the person to catch the ball

The person starts from the same point at the same instant and runs along the ground with constant speed $\dfrac{v_0}{2}$.

Since the person is on the ground, he can catch the ball only when the ball comes back to the ground, i.e. when $y=0$ again (other than $t=0$).

Time of flight of the projectile is:

$T = \frac{2v_0\sin\theta}{g}$

3. Horizontal range of the ball

The horizontal distance of the ball in this time is:

$R = v_0\cos\theta \cdot T$

Substitute $T$:

$R = v_0\cos\theta \cdot \frac{2v_0\sin\theta}{g}$ $R = \frac{2v_0^2\sin\theta\cos\theta}{g}$ $R = \frac{v_0^2\sin 2\theta}{g}$

4. Distance covered by the person in the same time

Runner's speed is:

$u = \frac{v_0}{2}$

So distance covered in time $T$ is:

$d = uT = \frac{v_0}{2}\cdot \frac{2v_0\sin\theta}{g}$ $d = \frac{v_0^2\sin\theta}{g}$

5. For catching the ball, distances must be equal

The person catches the ball when he reaches the landing point exactly at time $T$. So,

$d = R$

$\frac{v_0^2\sin\theta}{g} = \frac{v_0^2\sin 2\theta}{g}$

Cancel common terms:

$\sin\theta = \sin 2\theta$

Using $\sin 2\theta = 2\sin\theta\cos\theta$:

$\sin\theta = 2\sin\theta\cos\theta$

For nonzero projection angle, $\sin\theta \neq 0$, so:

$1 = 2\cos\theta$ $\cos\theta = \frac{1}{2}$ $\theta = 60^\circ$

6. Check the options

• A: No $\rightarrow$ incorrect
• B: Yes, $30^\circ$ $\rightarrow$ incorrect
• C: Yes, $60^\circ$ $\rightarrow$ correct
• D: Yes, $45^\circ$ $\rightarrow$ incorrect

Therefore, the person will be able to catch the ball if the projection angle is $60^\circ$.