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Question

For a given velocity of projection from a point on the inclined plane, the maximum range down the plane is three times the maximum range up the incline. Then, the angle of inclination of the inclined plane is

A
30
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B
45
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C
60
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D
90
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Solution

The correct option is A 30
Maximum range up the inclined plane (Rmax)up=u2g(1+sinα)
Maximum range down the inclined plane (Rmax)down=u2g(1sinα)
and according to problem : u2g(1sinα)=3×u2g(1+sinα)
By solving α=30

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