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Motion

Section: Physics

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Question : 11 of 12

Marks: +1, -0

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° 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

\mathrm{m/s^2}

[JEE Adv 2011 P2]

Your Answer:

Solution:

We have

s = ut +

\frac{1}{2} a t^2

Therefore, the vertical displacement is

S_v

= 0 = u sin 60° t +

\frac{1}{2} g t^2

⇒

\frac{10\sqrt{3}}{2} t - \frac{1}{2} 10 t^2

= 0

or t =

\sqrt{3}

s. The horizontal displacement is

S_H

= 1.15 m = ut cos 60° -

\frac{1}{2} a t^2

⇒ 1.15 =

5\sqrt{3} - \frac{a}{2} 3

Therefore, a =

\frac{2}{3} (5\sqrt{3} - 1.15)

= 5

\mathrm{m/s^2}

.

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