Question

A passenger in an open car travelling at $30\, m / s$ throws a ball out over the bonnet. Relative to the car the initial velocity of the ball is $20 m / s$ at $60^{\circ}$ 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

Answer: (B)

$\tan^{-1} \left(\frac{\sqrt{3}}{4}\right)$

Answer 2

$V _{ \text{car }}=30\, m'si$ $V _{ g \rightarrow \text{car }}=20 \cos 60^{\circ} \hat{1}+20 \sin 60^{\circ} \hat{1}$ $V _{ b \rightarrow \text{car }}=10 i +10 \sqrt{3} \hat{i}$ $\Rightarrow V _{ b \rightarrow \text { earth }}= V _{ b \rightarrow \text{car} }+ C _{\text {car } \rightarrow \text { earth }}$ $=10 \hat{i}+10 \sqrt{3} \hat{i}+(+30 \hat{i})$ $=+40 \hat{i}+10 \sqrt{3} \hat{i}$ $\Rightarrow$ Angle from horizontal $\theta=\tan ^{-1} \frac{y}{x}$ $\theta=\tan ^{-1} \frac{10 \sqrt{3}}{40}$ $\theta=\tan ^{-1} \frac{\sqrt{3}}{4} .$