Discussion Forum

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 m/s, at an angle of$60^{0}$ 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 in m/s², is

Answer:

Explanation:

 $t= T= \frac{2 u \sin \theta}{g}$

   =$\frac{2 \times 10 \times \sin 60^{0}}{10}= \sqrt{3} s$

Displacement of train in time  t = $\frac{1}{2}at^{2}$

Displacement of boy with respect to train = 1.15 m

$\therefore$ Displacement of boy with respect to ground = $(1.15+\frac{1}{2}at^{2})$

Displacement of ball with respect to  ground = $(u \cos 60^{0} ) t$

 To catch the ball back at initial height, 

 $1.15+\frac{1}{2} at^{2}= (u \cos 60^{0}) t$

 $\therefore$    $1.15+\frac{1}{2} a ( \sqrt{3})^{2}=10 \times \frac{1}{2} \times \sqrt{3}$

 Solving this equation , we get

 a=5 $ms^{-2}$

 $\therefore$ answer is 5