A body falls freely from some altitude H. At the moment the first body starts falling another body is thrown from the earth's surface which collides with the first at an altitude `h=H//2`. The horizontal distance is `l`. Find the initial velocity and the angle at which it was thrown.

Answer

`v_(0)=sqrt(gH(1+l^(1)/H^(2))), tan alpha=H/l`

Explanation

`H/2` distance covered by free falling body
`H/2=1/2 g t^(2), t=sqrt(H/g)`

In same time, projectile also travel vertical distance `H/2`, then `H/2=v sin alpha sqrt(H/2)-1/2gH/g`
`v sin alpha=sqrt(gH)` …(i)
also `l=v cos alpha sqrt(H/g), v cos alpha=lsqrt(g/H)` ...(ii)
From equation (i) and (ii)
`tan alpha=H/l v^(2) sin^(2) alpha+v^(2) cos^(2) alpha=gH+l^(2) g/H`
`v=sqrt(gH(1+l^(2)/H^(2))t)`