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 $10\,ms^{-1}$, at an angle of $60^°$ 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 of the train is

• A. $3\,ms^{-2}$
• B. $5\,ms^{-2}$
• C. $8\,ms^{-2}$
• D. $6\,ms^{-2}$

Answer 1

Answer: (B)

Here, $u= 10\,ms^{-1}$,
$\theta = 60^°$
$\therefore u\, cos\,\theta = 10 \times cos \,60^°$
$ = 5\,ms^{-1}$
and $u_{y}=u\,sin\,\theta=10\times sin\,60^{°}$
$=5\sqrt{3}\,ms^{-1}$
$\therefore t=\frac{2u_{y}}{g}$
$=\frac{2\times5\sqrt{3}}{10}$
$=\sqrt{3}\,s$
and $1.15=u_{x}t-\frac{1}{2}at^{2}$, where a is the acceleration of train
$1.15=5\times\sqrt{3}-\frac{1}{2}\times a\times\left(\sqrt{3}\right)^{2}$
$\frac{3a}{2}=5\sqrt{3}-1.15$
$=8.65-1.15=7.5$
$7.5\times\frac{2}{3}=5\,ms^{-2}$