A particle is projected under gravity with velocity 2ag from a point at a height h above the level plane at an angle θ to it. The maximum range R on the ground is:
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a
(a2+1)h
b
a2h
c
ah
d
2a(a+h)
answer is D.
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Detailed Solution
Coordinate of point P are (R, - h)Hence R ≤ 2a(a+h)-h = R tan θ- gR22(2ga)(1+tan2θ)or R2tan2θ-4aR tan θ+(R2-4ah) = 0For θ to be real.(4aR2) ≥ 4R2(R2-4ah)or 4a2 ≥ (R2-4ah)or R2 ≤ 4a(a+h)or R2 ≤ 2a(a+h) Rmax = 2a(a+h)