Question

A ball is thrown from the ground to clear a wall $3 m$ high at a distance of $6 m$ and falls $18 m$ away from the wall, the angle of projection of ball is

• A. $\tan ^{-1}\left(\frac{3}{2}\right)$
• B. $\tan ^{-1}\left(\frac{2}{3}\right)$
• C. $\tan ^{-1}\left(\frac{1}{2}\right)$
• D. $\tan ^{-1}\left(\frac{3}{4}\right)$

Answer 1

Answer: (B)

We know that range of a projectile
$R= u \frac{2 \sin 2 \theta}{g}$
Here $R=6+18=24$
$\therefore \frac{u^{2} \sin 2 \theta}{g}=24 \ldots \ldots (i)$
The equation trajectory of a projectile
$y=x\tan \theta -\frac{g{x}^2}{\mathrm{2u}{2}^{}{\cos }^2\theta }$
$3=6\tan \theta -\frac{36g}{\mathrm{2u}{2}^{}{\cos }^2\theta }$ ...... (ii)
From equation (i), $\frac{g}{u^2}=\frac{\sin 2\theta }{24}$
$=\frac{\sin \theta \cos \theta }{12}$
Substituting in equation (ii), we get
$3=6\tan \theta -\frac{3}{2}\tan \theta$
$3=\frac{9}{2}\tan \theta$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{2}{3}\right)$