Question

A ball is thrown from a point with a speed $V_0$ at an angle of projection '$\theta$'. From the same point and at the same instant a person starts running with a constant speed $\frac{V_0}{2}$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?

• A. $yes,{60}^0$
• B. $yes,{30}^0$
• C. $yes,{45}^0$
• D. $no$

Answer 1

The correct option is A $yes,{60}^0$

For person to catch the ball they should travel same distance along horizontal axis in time $T$.

Distance covered by person, $R=\frac{V_0}{2}\times T$....(i)

Time of flight of projectile, $T=\frac{2V_0\sin \theta }{g}$

$\Rightarrow R=\frac{V_0}{2}\times \frac{2V_0\sin \theta }{g}=\frac{2V_0^2\sin \theta }{2g}$....(ii)

Range of projectile, $R=\frac{V_0^2\sin 2\theta }{g}$...(iii)

Equating (ii) and (iii):

$\frac{V_0^2\sin 2\theta }{g}=\frac{2V_0^2\sin \theta }{2g}$

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

$\cos \theta =\frac{1}{2}$

$\theta ={60}^0$