Two balls A and B are projected at an angle of 45^ and 60^ respectively, so that the maximum heights reached are same for both. The ratio of initial velocity of projection of ball A to that for ball B is ( 30^ = 60^ = 1 2 , 45^ = 45^ = 1 2 , 60^ = 30^ = 3 2 )

Step-by-step solution

The maximum height reached by a projectile is given by H = u^2 ^2 2g For ball A, H_A = u_A^2 ^2 45^ 2g For ball B, H_B = u_B^2 ^2 60^ 2g Since H_A = H_B , we have: u_A^2 ^2 45^ 2g = u_B^2 ^2 60^ 2g u_A 45^ = u_B 60^ u_A u_B = 60^ 45^ Substituting the values 60^ = 3 2 and 45^ = 1 2 : u_A u_B = 3 2 1 2 = 3 2 The ratio of initial velocity of projection of ball A to that for ball B is 3 : 2 .