A boy playing on the roof of a `10 m` high building throws a ball with a speed of `10 m//s` at an angle of `30^(@)` with the horizontal. How far from the throwing point will the ball be at the height of `10 m` from the ground ?
`[ g = 10m//s^(2) , sin 30^(@) = (1)/(2) , cos 30^(@) = (sqrt(3))/(2)]`

Explanation

The ball will be at point P when it is a height of 10 m from the ground. So, we have to find distance OP, which can be calculated direct considering it is a projectile on a level OX at a height h from the horizontal level.

`OP = R = (u^(2) sin 2 theta)/(g) rArr R = (10^(2) xx sin (2 xx 30^(@)))/(10) = 8.66 m`