Question

A ball hit with a velocity $25 \,m / s$ at an angle $37^{\circ}$ with the horizontal just clears a wall horizontally. The ratio of height of the wall to its distance from the projection point of the ball is

• A. $\frac{3}{4}$
• B. $\frac{3}{2}$
• C. $\frac{3}{8}$
• D. $\frac{3}{6}$

Answer 1

Answer: (C)

Since the ball just clears the wall horizontally, the wall's height is equal to the maximum elevation of the ball and the wall is situated at a distance equal to half the range of the ball.
given $\theta=37^{\circ} ; R=\frac{u^{2} \sin 2 \theta}{g}$
and $H=\frac{u^{2} \sin ^{2} \theta}{2 g} $.
$\therefore \frac{H}{R}=\frac{\sin ^{2} \theta}{2 \sin 2 \theta}=\frac{1}{4} \tan \theta$
required ratio $=\frac{H}{R / 2}=\frac{2 H}{R}=2 \times\left(\frac{1}{4} \tan \theta\right)=\tan \theta$
$=\frac{1}{2} \tan 37^{\circ}=\frac{1}{2} \times 0.75=\frac{3}{8}$