A train is moving along a straight line with a constant acceleration $a$. A boy standing in the trai

Question

A train is moving along a straight line with a constant acceleration $a$. A boy standing in the train throws a ball forward with speed of $10 \mathrm{~m} \mathrm{~s}^{-1}$, at an angle of $60^{\circ}$ to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration (in $\mathrm{m} / \mathrm{s}^2$ ) of the train is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

(5) : Maximum height, $H_{\max }=\frac{u^2 \sin ^2 	heta}{2 g}$ \ Horizontal range, $R=\frac{u^2 \sin 2 	heta}{g}$
According to given problem $H_{	ext {max }}=\frac{R}{2}$
$
\begin{aligned}
& \frac{u^2 \sin ^2 	heta}{2 g}=\frac{u^2 \sin 2 	heta}{2 g} \
& \frac{u^2 \sin ^2 	heta}{2 g}=\frac{2 u^2 \sin 	heta \cos 	heta}{2 g} \
& \frac{\sin 	heta}{\cos 	heta}=2 \Rightarrow 	an 	heta=2 	herefore \sin 	heta=\frac{2}{\sqrt{5}} ; \cos 	heta=\frac{1}{\sqrt{5}} \
& 	herefore \quad R=\frac{2 u^2 \sin 	heta \cos 	heta}{g}=\frac{2 u^2 	imes \frac{2}{\sqrt{5}} 	imes \frac{1}{\sqrt{5}}}{g}=\frac{4 u^2}{5 g}
\end{aligned}
$