1.

A projectile is projected with a speed K Ve where Ve is escape velocity and K is a constant less than one. The maximum height reached by it from the centre of earth will be :

Answer»

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

Solution :According to conservation of energy
`(GMM)/R =(GMm)/(R+h)=1/2mK^(2)v_(e)^(2)`
or `gR- (gR^(2))/(R+h)=1/2K^(2)(sqrt(2gR))^(2)`
or `R/(R+h)=1-K^(2)`or `R+h=R/(1-K^(2))`


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