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A body is projected vertically upward from the surface of earth with a velocity sufficient to carry it to initially. Calculate the time taken by it to reach height `h`.

A

`sqrt((2R_(e))/(g))[(1+(h)/(R_(e)))^((3)/(2))-1]`

B

`(1)/(3)sqrt((R_(e))/(2g))[(1+(h)/(R_(e)))^(3//2)-1]`

C

`sqrt((R_(e))/(2g))[(1+(h)/(R_(e)))^((3)/(2))-1]`

D

`(1)/(3)sqrt((2R_(e))/(g))[(1+(h)/(R_(e)))^(3//2)-1]`

Text Solution

Verified by Experts

The correct Answer is:
D

Applying energy conservation from (1) to (2)
`(1)/(2)m.((2GM)/(R_(e)))-(GMm)/(R_(e))=(1)/(2)mv^(2)-(GMm)/(R_(e)+r)`
`implies (1)/(2)mv^(2)=(GMm)/(R_(e)+r)`
`implies v=sqrt((2GM)/(R_(e)+r))=(dr)/(dt)`
`implies sqrt(2GM)underset(0)overset(1)intdt=underset(0)overset(h)int(sqrt(R_(e)+r))dr`
`sqrt(2GM).t=(2)/(3)[(R_(e)+r)^(3//2)]_(0)^(h)`
`t=(2)/(3)sqrt((R_(e)^(3))/(2GM))[(1+(h)/(R_(e)))^(3//2)-1]`
`t=(1)/(3)sqrt((2R_(e))/(g))[(1+(h)/(R_(e)))^(3//2)-1]{because(GM)/(R_(e)^(2))=g}`
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