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 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`