A ball P is projected at an angle of 60^ with the vertical with certain initial speed. Another ball Q of the same mass as that of ball P is projected vertically upwards with the same initial speed as that of P. At the highest point, the ratio of potential energy of ball P to that of ball Q is ( 30^ = 0.5)

Step-by-step solution

Let the initial speed of both balls be u and their mass be m . For ball P, the angle of projection with the vertical is 60^ , so the angle with the horizontal is _P = 90^ - 60^ = 30^ . The maximum height reached by ball P is H_P = u^2 ^2 30^ 2g = u^2 (1/2)^2 2g = u^2 8g . The potential energy of ball P at the highest point is U_P = mgH_P = mg ( u^2 8g ) = mu^2 8 . For ball Q, it is projected vertically upwards, so the angle of projection with the horizontal is _Q = 90^ . The maximum height reached by ball Q is H_Q