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A stone is projected from point A with s...


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`

Text Solution

Verified by Experts

The correct Answer is:
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))`
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