A body is projected horizontally from the top of a building of height h. Velocity of projection is u. Find velocity with which the body will reach the ground.

Explanation

After a time t, the horizontal component of veclocity =u , and the vertical component of velocity=0+gt=gt.
`therefore` The resultant velocity at
`B=sqrt(u^2+("gt")^2) =sqrt(u^2+g^2(2h)/g)=sqrt(u^2+2gh)`
If it makes an angle `theta` with the horizontal , then
`theta =tan ^(-1) (("gt")/u)`