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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 maximum range Rmax is

A
Rmax=u2g(1sinβ)
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B
Rmax=u2g(1+sinβ)
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C
Rmax=ug(1sinβ)
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D
Rmax=ug(1+sinβ)
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Solution

The correct option is B Rmax=u2g(1+sinβ)
The expression of horizontal range R=2u2sin(αβ)cosαgcos2β

From Identity 2sinAcosB=sin(A+B)+sin(AB)
2sin(αβ)cosβ=sin(2αβ)+sin(β)

R=u2gcos2β[sin(2αβ)sinβ]

Range R along incline is maximum if 2αβ=π/2

so, Rmax=u2gcos2β[1sinβ]

=u2g(1sin2β)[1sinβ)]

=u2g(1+sinβ)(1sinβ)[1sinβ]=u2g(1+sinβ)

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