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A particle is released from height S from the surface of the earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the earth and the speed of the particle at that instant are respectively:


 

Option: 1

\frac{S}{4},\frac{3gS}{2}
 


Option: 2

\frac{S}{4},\frac{\sqrt{3gS}}{2}


Option: 3

\frac{S}{2},\frac{\sqrt{3gS}}{2}

 


Option: 4

\frac{S}{4},\sqrt{\frac{{3gS}}{2}}


Answers (1)

best_answer

1\rightarrow Initial

                                                          2\rightarrow Final

TE_{1}=TE_{2}

mgs+0=\left ( mgh \right )+KE

\left ( kE \right )=3\left ( mgh \right )\left ( Given \right )

mgs=4mgh

h=\frac{s}{4}

3mgh=\frac{1}{2}mv^{2}

v=\sqrt{2gh\times 3}=\sqrt{2g\frac{s}{4}\times 3}=\sqrt{\frac{gs}{2}\times 3}

v=\sqrt{\frac{3gS}{2}}

Posted by

jitender.kumar

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