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Question

# If R1 and R2 are the maximum ranges of a projectile when it is projected from the bottom and top respectively of an inclined plane respectively, then find the maximum range of the projectile when it is projected from the ground with same velocity.

A
R1+R22
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B
R1R2R1+R2
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C
2R1R2R1+R2
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D
R1+R2R1R2
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Solution

## The correct option is C 2R1R2R1+R2When the projectile is from top to bottom of the inclined plane. Resolving the motion parallel and perpendicular to the inclined plane as shown in the diagram, vx=vcos(α−θ)vy=vsin(α−θ)ax=−gsinθay=−gcosθ If T is the time of flight, then 0=vsin(α−θ)T−12gcosθ T2 (Using second equation of motion along Y-axis) ⇒T=2vsin(α−θ)gcosθ(1) If the projected object covers horizontal distance OB in time T, OB=vcosα×T Now, cosθ=OBOA (From the figure) ⇒OA=OBcosθ=vcosα×Tcosθ Using (1), ⇒OA=v2gcos2θ[2sin(α−θ)cosα] Using the identity 2sinAcosB=sin(A+B)+sin(A−B) OA=R1=v2gcos2θ[sin(2α−θ)−sinθ] Clearly, the range R1(=OA) will be maximum when sin(2α−θ) is maximum, i.e., 1. This would mean 2α−θ=π2 or α=θ2+π4 Maximum range on the inclined plane from bottom to top can be denoted by R1. ∴R1=v2gcos2θ(1−sinθ)=v2(1−sinθ)g(1−sin2θ)=v2(1−sinθ)g(1+sinθ)(1−sinθ)=v2g(1+sinθ) When the projectile is from top to bottom of the inclined plane. Range of a projectile when it is from bottom of an inclined plane was found to be OA=v2gcos2θ[sin(2α−θ)−sinθ] Here α is the angle made by the direction of projection with the horizontal when it is projected from the bottom of the plane. When it is projected from the top of the plane, the angle made by the projectile with the inclined plane is α+θ. That is θ is replaced with −θ in the above formula. Hence, OA=v2gcos2θ[sin(2a−(−θ))−sin(−θ)]=v2gcos2θ[sin(2α+θ)+sinθ] OA is maximum when 2α+θ=π2 and sin(2α+θ)=1 ⇒OAmax=v2gcos2θ[1+sinθ]⇒OAmax=R2=v2g(1−sin2θ)[1+sinθ]R2=v2g(1−sinθ) We know that the maximum range of a projectile from the bottom and top of an inclined plane are, respectively R1=v2g(1+sinθ) and R2=v2g(1−sinθ) Then, 1R1+1R2=g(1+sinθ)v2+g(1−sinθ)v2 ⇒R1+R2R1R2=2gv2⇒2R1R2R1+R2=v2g Maximum range of a projectile on a horizontal surface Rmax=v2g Thus, Rmax=2R1R2R1+R2

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