A projectile is thrown from the ground such that it reaches a maximum height H above the ground and reaches a horizontal range R. The area between the path of the projectile and horizontal ground in the plane of its motion is

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RH3

RH2

2RH3

3RH2

answer is C.

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Detailed Solution

Equation of trajectory can be written as y=x tanθ1−xR but tanθ=4HR hence y=4HRx1−xR Required area=∫ydx=∫0R4HRx−x2Rdx=4HRR22−R23=23RH

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