Question

A train is moving along a straight line with a constant acceleration. A boy standing inside the train throws a ball in forward direction with a speed of $10\text{m/s}$ at an angle of ${60}^{\circ }$ to the horizontal, with respect to himself. The boy has to move forward by $1.15\text{m}$ inside the train to catch the ball (at the initial height). Then,
[Take $g=10{\text{m/s}}^2$]

• A. Acceleration of train is $5{\text{m/s}}^2$
• B. Acceleration of ball is $10{\text{m/s}}^2$
• C. Acceleration of ball is $5\sqrt{5}{\text{m/s}}^2$
• D. Displacement of train is $8.66\text{m}$

Answer 1

Answer: (A), (B)

The correct options are
A Acceleration of train is $5{\text{m/s}}^2$
B Acceleration of ball is $10{\text{m/s}}^2$

$u=10\text{m/s}$
Angle of projection, $\theta ={60}^{\circ }$
Distance covered by boy to catch the ball at the same height is, $d=1.15\text{m}$
Acceleration of train is $a$ and starts from rest when ball is thrown.


Since the only force acting on the ball is $mg$ downwards, so its motion will be that of a projectile with $u=10\text{m/s}$. Acceleration of ball will be $g=10{\text{m/s}}^2$ downwards.
$\therefore$ Horizontal range $R=\frac{u^2\sin 2\theta }{g}$
$=\frac{{10}^2\sin {120}^{\circ }}{10}=\frac{100\times \left(\frac{\sqrt{3}}{2}\right)}{10}=5\sqrt{3}\text{m}$
Time of flight of ball, $T=\frac{2u\sin \theta }{g}=\frac{2\times 10\times \sin {60}^{\circ }}{10}$
$\therefore T=\sqrt{3}\text{s}$
During time interval $T$, displacement of train $S=uT+\frac{1}{2}a{T}^2$
$\Rightarrow S=\frac{1}{2}a\times 3$
[$\because$ train starts from rest when ball is thrown]
Boy needs to move by $1.15\text{m}$ w.r.t train, to catch the ball:
$\Rightarrow \frac{3}{2}a+1.15=\text{Range of ball}$
$\Rightarrow \frac{3a}{2}+1.15=5\sqrt{3}$
$\Rightarrow a=\frac{(5\sqrt{3}-1.15)\times 2}{3}\simeq 5{\text{m/s}}^2$
So, Displacement of train
$S=\frac{3a}{2}\simeq 7.5\text{m}$
Option A and B are correct.