A small stone is tied to a string of length 2 m and whirled in a vertical circle. At the lowest point, its speed is 7gL . When the stone is ascending and the string makes an angle of 60^ with the upward vertical, the string is suddenly cut. The stone then moves freely under gravity as a projectile. The maximum height reached by the stone from the lowest point of the circular path is : (Take acceleration due to gravit

Step-by-step solution

Let the lowest point of the circular path be the reference level ( h = 0 ). The initial speed at the lowest point is u = 7gL . When the string makes an angle of 60^ with the upward vertical, the angle it makes with the downward vertical is 180^ - 60^ = 120^ . The height of the stone at the instant the string is cut is: h_c = L - L (120^ ) = L - L (- 1 2 ) = 3L 2 Substituting L = 2 m , we get h_c = 3 m . By conservation of mechanical energy, the speed v_c of the stone at this point is: 1 2 mu^2 = 1 2 mv_c^2 + mgh_c