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. `(u t)/(sqrt(3))`
B. `(sqrt(3) ut)/(2)`
C. `sqrt(3) u t`
D. `2 ut`

Answer 1

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