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Question

A plane surface is inclined making an angle θ with the horizontal. From the bottom of this inclined plane, a bullet is fired with velocity v. The maximum possible range of the bullet on the inclined plane is

A
v2g
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B
v2g(1+sinθ)
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C
v2g(1sinθ)
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D
v2g(1+cosθ)
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Solution

The correct option is B v2g(1+sinθ)

Resolving the motion parallel and perpendicular to the inclined plane as shown in the diagram. So from the diagram,

vx=vcos(αθ); vy=vsin(αθ)
ax=gsinθ; ay=gcosθ

If T is the time of flight, then applying second equation of motion along yaxis.

y=vyt+12ayt2

0=vsin(αθ)T12gcosθ T2

T=2vsin(αθ)gcosθ ..........(1)

If the bullet covers horizontal distance OB in time T,

OB=vcosα×T

Now,
cosθ=OBOA (From the figure)

OA=OBcosθ=vcosα×Tcosθ

Using equation (1),

OA=v2gcos2θ[2sin(αθ)cosα]

Using the identity,
2sinAcosB=sin(A+B)+sin(AB)

OA=v2gcos2θ[sin(2αθ)sinθ]

Clearly, the range R(=OA) will be maximum when sin(2αθ) is maximum, i.e., 1.

This would mean
2αθ=π2 or α=θ2+π4

Maximum range on the inclined plane can be denoted by Rmax.

Rmax=v2gcos2θ(1sinθ)

Rmax=v2(1sinθ)g(1sin2θ)=v2(1sinθ)g(1+sinθ)(1sinθ)

Rmax=v2g(1+sinθ)

Hence, option (B) is correct.

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