Correct Answer - `2sqrt(2)h`
Since `y=gx^(2)//2v^2` for a horizontal projection for the same `x` we have `y_(1),y_(2)` and `v_(1), v_(2)`.Putting `y_(1)-y_(2)=h` and other values find `x` and `y_(s)`.
`y_(1)=(gx^(2))/(2.2gh)` and `y_(2)=(gx^(2))/(2.4gh)`
`y_(1)-y_(2)=h=(gx^(2))/(8gh)impliesx=2sqrt(2h)`
and `y_(2)=(g(8h^(2)))/(2.4.gh)implies y_(2)=h`
