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(1−sinθ)
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D
v2g(1+cosθ)
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Solution
The correct option is Bv2g(1+sinθ)
Resolving the motion parallel and perpendicular to the inclined plane as shown in the diagram. So from the diagram,
vx=vcos(α−θ);vy=vsin(α−θ) ax=−gsinθ;ay=−gcosθ
If T is the time of flight, then applying second equation of motion along y−axis.
y=vyt+12ayt2
0=vsin(α−θ)T−12gcosθT2
T=2vsin(α−θ)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(α−θ)cosα]
Using the identity, 2sinAcosB=sin(A+B)+sin(A−B)
OA=v2gcos2θ[sin(2α−θ)−sinθ]
Clearly, the range R(=OA) will be maximum when sin(2α−θ) is maximum, i.e., 1.
This would mean 2α−θ=π2orα=θ2+π4
Maximum range on the inclined plane can be denoted by Rmax.