A stone is thrown by a student from the bottom of a hill with a velocity `30 ms^(-1)` making an angle of `60^(@)` with the horizontal. If the slpe of the hill is `30 ^(@)` with the horizontal. Find the distance from the student to a point at which the stone fals on hill, use `g= 10 ms^(-2)`.

Explanation

Refr to Art. 2 (d) .9, the distance
coverd on slope is `R= (2 u^(2) cos theta sin (theta-theta_(0)))/(g cos^(2) theta_(0))`
Here, `u =30 ms^(-1)`
`theta =60^(@), theta =30^(@), g=10 ms^(-2)`
:. `R =(2 xx 30^(2) cos 60^(@) sin 9 60^(@) -30^(@)))/(10 xx cos ^(2) 30(#))`
`=(2 xx 900 xx (1//2) xx (1//2))/(10 xx (sqrt3//2)^(2)`
`=69 m`.