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, with respect to the train 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:
5
4
1
6
Solution:
Ball performs parabolic motion, For vertical motion,s=ut - ½gt² ∴0=10sin60°t - ½×10×t² ∴t=√3s For horizontal motion,s=ut+½at² ∴1.15=10cos60°×√3 + ½a(3) ∴a=5m/s²