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A particle is projected from the bottom of an inclined plane of inclination `30^@`. At what angle `alpha `(from the horizontal) should the particle be projected to get the maximum range on the inclined plane.

A

`45^(@)`

B

`53^(@)`

C

`76^(@)`

D

`60^(@)`

Text Solution

Verified by Experts

The correct Answer is:
D
For maximum range:

`alpha=pi/4-( theta_(0))/2=45^(@)-(30^(@))/2=30^(@)`
Angle with horizontal:
`theta=alpha+ theta_(0)=30^(@)+30^(@)=60^(@)`
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