1.

A projectile is fired from earth vertically with a velocity Kv_(e ) where v_(e )= escape velocity and K is a constantless than unity. What is the maximum height to which it rises as measured from centre of earth?

Answer»

`(R )/(K^(2))`
`(R )/(1-K^(2))`
`(R )/(K^(2)-1)`
`(K^(2)R)/(R-1)`

Solution :Let r be the maximum distance from the CENTRE of EARTH to which projectile rises, then,
`(1)/(2)mk^(2)v_(e )^(2)=(GMm)/(R )-(GMm)/(r )`
`(1)/(2) mk^(2)(2GM)/(R )=GMm[(1)/(R )-(1)/(r )]`
`(1)/(r )=(1)/(R )-(k^(2))/(R )=(1-k^(2))/(R )`
or`r=(R )/(1-k^(2))`.
Correct choice is (b).


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