A train is moving along a straight line with a constant acceleration a. A boy standing in the train throws a…
A train is moving along a straight line with a constant acceleration a. A boy standing in the train throws a ball forward with a speed of $10 \mathrm{~m} / \mathrm{s}$, at an angle of $60^{\circ}$ to the horizontal. The boy has to move forward by $1.15 \mathrm{~m}$ inside the train to catch the ball back at the initial height. The acceleration of the train in $\mathrm{m} / \mathrm{s}^2$, is
Solution
$\begin{aligned} t & =T=\frac{2 u \sin \theta}{g} \\ & =\frac{2 \times 10 \times \sin 60^{\circ}}{10}=\sqrt{3} \mathrm{~s}\end{aligned}$
Displacement of train in time $t=\frac{1}{2} a t^2$
Displacement of boy with respect to train
$
=1.15 \mathrm{~m}
$
$\therefore$ Displacement of boy with respect to ground $=\left(1.15+\frac{1}{2} a t^2\right)$
Displacement of ball with respect to ground $=\left(u \cos 60^{\circ}\right) t$
To catch the ball back at initial height,
$
\begin{aligned}
1.15+\frac{1}{2} a t^2 & =\left(u \cos 60^{\circ}\right) t \\
\therefore 1.15+\frac{1}{2} a(\sqrt{3})^2 & =10 \times \frac{1}{2} \times \sqrt{3}
\end{aligned}
$
Solving this equation, we get
$
a=5 \mathrm{~ms}^{-2}
$
$\therefore$ Answer is 5 .
Analysis of Question
(i) Question is moderately tough.
(ii) Velocity of ball given in the question is with respect to ground.
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