Question

A passenger in an open car travelling at $30m/s$ throws a ball out over the bonnet. Relative to the car the initial velocity of the ball is $20m/s$ at ${60}^0$ to the horizontal. The angle of projection of the ball with respect to the horizontal road will be :

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

Answer 1

The correct option is B ${\tan }^{-1}\left(\frac{\sqrt{3}}{4}\right)$

$V_{car}=30m^{\prime }s\hat{i}$

$V_{g\to car}=20\cos {60}^o\hat{i}+20\sin {60}^o\hat{i}$

$V_{b\to car}=10\hat{i}+10\sqrt{3}\hat{i}$

$\Rightarrow V_{b\to earth}=V_{b\to car}+C_{car\to earth}$

$=10\hat{i}+10\sqrt{3}\hat{i}+(+30\hat{i})$

$=+40\hat{i}+10\sqrt{3}\hat{i}$

$\Rightarrow$ Angle from horizontal

$\theta ={\tan }^{-1}\frac{y}{x}$

$\theta ={\tan }^{-1}\frac{10\sqrt{3}}{40}$

$\theta ={\tan }^{-1}\frac{\sqrt{3}}{4}$.