Question

A train is moving along a straight line with a constant acceleration $a$. A boy standing in the train throws a ball upwards with a speed of $10\text{m/s}$ at an angle ${30}^{\circ }$ to the horizontal. The boy has to move forward by $5\text{m}$ inside the train to catch the ball back at its initial height. The acceleration of the train in ${\text{m/s}}^2$ is:

• A. $1.2$
• B. $2.3$
• C. $3.3$
• D. $7.3$

Answer 1

The correct option is D $7.3$

Given that,
Initial speed of ball w.r.t ground $\left(u\right)=10\text{m/s}$
Angle of projectile, $\theta ={30}^{\circ }$
Distance covered by boy w.r.t train $=5\text{m}$
Time of flight of ball is
$T=\frac{2u\sin \theta }{g}$
$\Rightarrow T=\frac{2\times 10\times \sin {30}^{\circ }}{10}$
$\Rightarrow T=1\text{sec}$

Displacement of the train in time $t=\frac{1}{2}a{t}^2$
Displacement of boy w.r.t train $=5\text{m}$
Displacement of boy w.r.t ground $=(5+\frac{1}{2}a{t}^2)$
Displacement of ball w.r.t ground $=\left(u\cos {30}^{\circ }\right)t$
To catch the ball back at initial height,
$5+\frac{1}{2}a{t}^2=u\cos {30}^{\circ }t$
$\Rightarrow 5+\frac{1}{2}\times a{\left(1\right)}^2=10\times \frac{\sqrt{3}}{2}\times 1$
$\Rightarrow a=(5\sqrt{3}-5)\times 2$
$\Rightarrow a=7.32{\text{m/s}}^2$