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 $10m{s}^{-1}$, at an angle of $60^{°}$ to the horizontal. The boy has to move forward by $1.15m$ inside the train to catch the ball back at the initial height. The acceleration of the train is

• A. $3m{s}^{-2}$
• B. $5m{s}^{-2}$
• C. $8m{s}^{-2}$
• D. $6m{s}^{-2}$

Answer 1

Answer: (B)

Here, $u=10m{s}^{-1}$,
$\theta =60^{°}$
$\therefore ucos\theta =10\times cos60^{°}$
$=5m{s}^{-1}$
and $u_y=usin\theta =10\times sin60^{°}$
$=5\sqrt{3}m{s}^{-1}$
$\therefore t=\frac{2u_y}{g}$
$=\frac{2\times 5\sqrt{3}}{10}$
$=\sqrt{3}s$
and $1.15=u_{x}t-\frac{1}{2}a{t}^2$, where a is the acceleration of train
$1.15=5\times \sqrt{3}-\frac{1}{2}\times a\times {\left(\sqrt{3}\right)}^2$
$\frac{3a}{2}=5\sqrt{3}-1.15$
$=8.65-1.15=7.5$
$7.5\times \frac{2}{3}=5m{s}^{-2}$