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Question

An inclined plane is making an angle β with horizontal. A projectile is projected from the bottom of the plane with a speed u at an angle α with horizontal then its range R when it strikes the inclined plane is :

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A
2u2sin(αβ)cosαgcos2β
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B
u2sin(αβ)cosαgcos2β
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C
2u2sin(α+β)cosαgcos2β
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D
u2sin(α+β)cosαgcos2β
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Solution

The correct option is A 2u2sin(αβ)cosαgcos2β
To calculate time of flight:
We know that the total displacement perpendicular to the inclined plane is zero.
So, 0=usin(αβ)t12gcosβt2
t=2usin(αβ)gcosβ
Now, for range:
OB=(ucosα)t=2u2sin(αβ)cosαgcosβ
OA=OB/cosβ=2u2sin(αβ)cosαgcos2β=R

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