A ball of mass (m) is thrown vertically up. Another ball of mass ` 2 m` is thrown at an angle ` theta` with the vertical. Both of them stay in air for the same period of time. What is the ratio of the height attained by two balls.

Explanation

For the ball thrown vertically upwards, the time taken by the ball to come back is `T_(1)=(2u_(1))/(g)`
For the ball projected at an angle `theta` with their vertical, the time of flight is `T_(2)=2u_(2) cos theta/g`
Since time of flights for the balls is same, so `(2u_(1))/(g)=(2u_(2)cos theta)/(g)` or `u_(1)=u_(2) cos theta`
`because h_(1)=(u_(1)^(2))/(2g)` and `h_(2)=(u_(2)^(2))/(2g) cos ^(2) theta therefore (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`