Question

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

The correct option is B ${\tan }^{-1}\left(\frac{2}{3}\right)$

We know that range of a projectile
$R=\frac{U^2\sin 2\theta }{g}$
Here $R=6+18=24$
$\therefore \frac{U^2\sin 2\theta }{g}=24$ .....(i)
The equation of trajectory of a projectile
$y=x\tan \theta -\frac{g{x}^2}{2U^2{\cos }^2\theta }$
$3=6\tan \theta -\frac{36g}{2U^2{\cos }^2\theta }$ ....(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
$3=6\tan \theta -\frac{3}{2}\tan \theta$
$3=\frac{9}{2}\tan \theta$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{2}{3}\right)$