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

Answer: (B)

We know that range of a projectile
$R=\frac{u^{2} sin \left(2 \theta \right)}{g}$
Here $R=6+18=24$
$\therefore \, \, \frac{u^{2} sin 2 \theta }{g}=24$ ...... (i)

The equation trajectory of a projectile
$y=xtan \theta -\frac{g x^{2}}{2u^{2} cos^{2} ⁡ \theta }$
$3=6tan \theta -\frac{36 g}{2u^{2} cos^{2} ⁡ \theta }$ ...... (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=6tan \theta -\frac{3}{2}tan⁡\theta $
$3=\frac{9}{2}tan \theta $
$\Rightarrow \, \, \, \theta =\left(tan\right)^{- 1} \left(\frac{2}{3}\right)$