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. .......


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.