The surface of a hill is inclined at an angle `alpha` to the horizontal (Fig.). A stone is thrown from the summit of the hill at an initial speed `v_(0)` at an angle `beta` to the vertical. How far from the summit will the stone strike the ground?

Answer

`b=(2v_(0)^(2)sin beta cos (alpha-beta))/(g cos^(2)alpha)`

Explanation

The law of motion of the stone is
`y=h +v_(0)t cos beta` -
`-(1)/(2) g t^(2) , x=v_(0)t sin beta`
When the stone strikes the hill,
`x_(B) = b cos alpha, y_(B) = h - b sin alpha`
Substituting this into the law of motion we obtain the result sought.