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A particle of mass 'm' is projected with a velocity v=KV_{e}\left ( K< 1 \right )from the surface of the earth .\left ( V_{e}=escape velocity \right )

The maximum height above the surface reached by the particle is :    

Option: 1

R\left ( \frac{K}{1-K} \right )^{2}


Option: 2

R\left ( \frac{K}{1+K} \right )^{2}


Option: 3

\frac{R^{2}K}{1+K}


Option: 4

\frac{RK^{2}}{1-K^{2}}


Answers (1)

best_answer

V=kv_{e}\left ( k< 1 \right )

TE_{1}=TE_{2}

\frac{1}{2}mv^{2}+\left ( -\frac{GMm}{R} \right )=0+\left ( \frac{-GMm}{R+h} \right )

\frac{1}{2}mxk^{2}\times \frac{2GM}{R}-\frac{GMM}{R}=\frac{-GMm}{R+h}

\frac{GMm}{R}\left [ K^{2}-1 \right ]=\frac{-GMm}{R+h}

\left [ 1-k^{2} \right ]=\frac{R}{R+h}

\frac{1}{\left ( 1-k^{2} \right )}=1+\frac{h}{R}

\frac{k^{2}R}{\left ( 1-k^{2} \right )}=h

Posted by

Ritika Harsh

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