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 $R=6+18=24$
$\therefore \frac{u^{2}sin2\theta }{g}=24$ ...... (i)

The equation trajectory of a projectile
$y=xtan\theta -\frac{g{x}^2}{2u^{2}co{s}^2\theta }$
$3=6tan\theta -\frac{36g}{2u^{2}co{s}^2\theta }$ ...... (ii)
From equation (i), $\frac{g}{u^2}=\frac{sin2\theta }{24}$
$=\frac{sin\theta cos\theta }{12}$
Substituting in equation (ii), we get
$3=6tan\theta -\frac{3}{2}tan\theta$
$3=\frac{9}{2}tan\theta$
$\Rightarrow \theta ={\left(tan\right)}^{-1}\left(\frac{2}{3}\right)$