A ball of mass M is thrown vertically upwards. Another ball of mass 2M is thrown at an angle `theta` with the vertical. Both of them stay in air for the same period of time. The heights attained by the two are in the ratio

Explanation

For the ball thrown vetically upwards, the time taken by the ball to come back is,
` T_(1) =2 u_(1)//g`.
For the ball projected at angle ` theta` with their vertical, the time of flight is,
` T_(2) =2 u_(2) cos theta //g`
since time for both the balls is same, so
` (2 u_1)/g = (2 u_2 cos theta)/g or u_1 =u_2 cos theta`.
Now, ` h_1 = u_1^2/(2 g) and h_2 = u_2^2/ (2 g) cos ^2 theta ,
Hence, ` h_1 /h_2 = u_1 ^2 /(u_2 ^2 cos ^2 theta) = (u_2^2 cos&2 theta)/(u_2^2 cos^2 theta) =1`.