Two stones are thrown vertically upwards simultaneously with their initial velocities `u_(1) and u_(2)` respectively. Prove that the heights reached by them would be in the ratio of `u_(1)^(2) : u_(2)^(2)` (Assume upward acceleration is `-g` and downward acceleration to be ` + g`).

Explanation

At the highest point, `v = 0`.
`:.` From `v^(2) - u^(2) = 2 gh`
`0 - u^(2) = 2(-g)h` , `h = (u^(2))/(2 g)`
For two stones : `h_(1) = (u_(1)^(2))/(2g) and h_(2) = (u_(2)^(2))/(2g)`
`:. (h_(1))/(h_(2)) = (u_(1)^(2)//2g)/(u_(2)^(2)//2g) = (u_(1)^(2))/ (u_(2)^(2))`
which was to be proved.