Question

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

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


Using equation of projectile

$y=x\tan \theta (1-\frac{x}{R}).....\left(1\right)$

From diagram,

Range, $R=6+18=24\text{m}$ and

For point $\text{P},x=6\text{m},y=3\text{m}$

putting the all value in eq. $\left(1\right)$, we get

$3=6\tan \theta (1-\frac{6}{24})$

$\Rightarrow \tan \theta =\frac{2}{3}$

$\Rightarrow \theta ={\tan }^{-1}\left(\frac{2}{3}\right)$

Therefore, option $\left(b\right)$ is correct.