## Kinematics

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 m/s2, is. .......

(2011)

Solution : Acceleration of the ball with respect to the train is $$\leftarrow a$$ and $$\downarrow g$$. The total time of flight is subjected to vertical acceleration g and given by $$T = \frac{2usin\theta}{g}$$. where u = 10 m/s and $$\theta = 60^{0}$$. Hence T = $$\sqrt{3}$$ s. Now using the fact that the horizontal displacement of the ball with respect to train is 1.15 m. Using $$S = ut + \frac{1}{2}at^{2}$$, we write, $$1.15 = 10 cos 60^{0}\sqrt{3} - \frac{1}{2}a(\sqrt{3})^{2}$$. Solving, we get, a = 5m/s2.