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

For lst stone,given initial velocity u = `u_(1)`
Let height attained by it be `h_(1)`
Form third equation of motion, `v^(2)=u^(2)` - 2gh for upward motion
[Here, we have used nugative sigh, because as given upward acceleration is taken to be -q
At the highest point the velocity becomes zero i.e., v = 0
`0 = u_(1)^(2)-2gh_(1)`
`rArr 2gh_(1)= u_(1)^(2) rArr h_(1)= (u_(1)^(2))/(2g)`
Similarly,
For llnd stone, given initial velocty, u = `u_(2)`
Let the height attained by it be `h_(2)`.
From third equation of motion, `v^(2) = u^(2) - 2gh`
[at the highest point the velocity becomes zero i.e., v = 0]
`0=u_(2)^(2)-2gh_(2)`
`:.2gh_(2)=u_(2)^(2)`
`rArr h_(2)= (u_(2)^(2))/(2g)`
Now , ratio of heights reached by the two stones
`h_(1):h_(2)= (u_(1)^(2))/(2g):(u_(2)^(2))/(2g)`
`:. h_(1):h_(2)=u_(1)^(2):u_(2)^(2)` Hence proved