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 :
A
2u2sin(α−β)cosαgcos2β
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B
u2sin(α−β)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 A2u2sin(α−β)cosαgcos2β To calculate time of flight:
We know that the total displacement perpendicular to the inclined plane is zero. So, 0=usin(α−β)t−12gcosβt2 ∴t=2usin(α−β)gcosβ Now, for range: OB=(ucosα)t=2u2sin(α−β)cosαgcosβ ∴OA=OB/cosβ=2u2sin(α−β)cosαgcos2β=R