If R is the range of a projectile on a horizontal plane and h its maximum height, the maximum horizontal range with the same velocity of Projection is:
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a
2h
b
R2 /8h
c
2R+h28R
d
2h+R28R
answer is D.
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Detailed Solution
R=2u2sinθcosθ2gand h=u2sin2θ2g⇒R2=4u4sinθ2cosθ2g2R28h=u2g−2h⇒maximum range = u2g=R28h+2h
If R is the range of a projectile on a horizontal plane and h its maximum height, the maximum horizontal range with the same velocity of Projection is: