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# If a body is to be projected vertically upwards from the surface of the earth to reach a height nR then the velocity with which it is to be projected is

$(a)\;\sqrt {\frac{2ngR}{n+1}} \quad (b)\;\sqrt {\frac{ngR}{n-1}} \quad (c)\;\sqrt {\frac{2ngR}{n}} \quad (d)\;\sqrt {\frac{2ngR}{n-1}}$

Decrease in Kinetic energy = increase in gravitational potential energy
$\large\frac{1}{2}$$mv^2= \Delta v \large\frac{1}{2}$$ mv^2=\large\frac{mgh}{1+\Large\frac{h}{R}}$ where h is the height above earth
$\large\frac{1}{2}$$mv^2=\large\frac{mgnR}{1+\Large\frac{nR}{R}}$
Solving we get,
$v^2= \large\frac{2gnR}{n+1}$
$v= \sqrt {\large\frac{2 ngR}{n+1}}$
Hence a is the correct answer.

edited Feb 18, 2014 by meena.p