Kalpit and Mukesh are playing with two different balls of masses m and 2m respectively. If Kalpit throws his ball vertically up and Mukesh at an angle θ,such that time of flight of both balls is same . The height attained by the two balls are in the ratio of :

Let u1 and u2 be the initial velocities respectively. If h1 and h2 are theheights attained by them, thenh1 =u122g  and   h2 =u22sin2θ2g                        ...1The times of ascent of balls are equal, we have tt=u1/g =u2 sin θ/g∴u1 = u2 sinθ                                                  ... 2divide the heights of equation (1)From eq.  1  h1h2=u12u22sin2θ                       ... 3substitute equation(2) in (3) h1h2=11