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A ball of mass m falls vertically to the ground from a height h1 and rebound to a height h2. The change in momentum of the ball on striking the ground is 

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a
mg(h1−h2)
b
m(2gh1+2gh2)
c
m2g(h1+h2)
d
m2g(h1+h2)

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detailed solution

Correct option is B

When ball falls vertically downward from height h1 its velocity v→1=2gh1and its velocity after collision  v→2=2gh2Change in momentumΔP→=m(v→2−v→1)=m(2gh1+2gh2)(because v→1 and v→2 are opposite in direction)

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