The surface of a hill is inclined at an angle `alpha` to the horizontal .A stone is thrown from the summit of the hill at an angle `beta` to the vertical with velocity `v_0`.How far from the summit will the stone strike the ground ?

Explanation

Take the axis as shown in the figure
`U_x=V_0sin(beta-alpha)`
`U_y=V_0cos(beta-alpha)`
`a_x=g sinalpha`
`a_y=-gcosalpha`
Applying kinematic equation along the X and the Y-axis
`x=V_0sin(beta-alpha)t+(1)/(2)g(sinalpha)t^(2)`….(1)
and `0=v_0cos(beta-alpha)t-(1)/(2)g(cosalpha)t^(2)....(2)`
Soloving (1) and(2)we get
`t=(2V_0 cos(beta-alpha))/(gcosalpha)`
So, `x=V_0 sin(beta-alpha)(2V_0cos(beta-alpha))/(gcosalpha)+(1)/(2)gsinalpha(4v_(0)^(2)cos^(2)(beta-alpha))/(g^(2)cos^(2)alpha)`
`x=(2V_(0)^(2)cos(beta-alpha))/(gcosalpha){sin(beta-alpha)cosalpha+cos(beta-alpha)sinalpha}`
`(2V_(0)^(2)cos(beta-alpha))/(gcos^(2)alpha)`