A body falls freely from some altitude H (Fig.). 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^(2)//H^(2))); tan alpha = H//l`.

Explanation

The laws of motion of the first and of the second objects are
`x_(1)=l, y_(1)=H-(1)/(2) g t^(2)`
`x_(2)=v_(0)t cos alpha , y_(2)= v_(0) t sin alpha - (1)/(2) g t^(2)`
When they meet, `x=l, y=H//2`. It follows that
`l=v_(0) t cos alpha, (H)/(2) = v_(0) t sinalpha - (1)/(2) g t^(2), (H)/(2)=H -(1)/(2)g t^(2)`