Question

A stone is projected from point A with speed u making an angle `60^@` with horizontal as shown. The fixed inclined suface makes an angle `30^@` with horizontal. The stone lands at B after time t. Then the distance AB is equal to
A. `(ut)/(sqrt(3))`
B. `(sqrt(3)ut)/(2)`
C. `sqrt(3)ut`
D. `2 ut`

Answer 1

Correct Answer - A
The horizontal displacement in time `t` is
`AC=u cos 60^(@)t=(ut)/(2)`
`:. ` Range on inclined plane `=(AC)/(cos 30)=(ut)/(sqrt(3))`