Kelvi

A particle of mass $m$ is projected with a velocity $v = k V_e (k < 1)$ from the surface of the Earth. $\newline (V_e =$ escape velocity $) \newline$ The maximum height above the surface reached by the particle is:

hard

Gravitation

2021

physics

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

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

Explanation

To solve this problem, we need to determine the maximum height reached by a particle projected with a velocity $v = k V_e$ from the surface of the Earth. $\\$ The escape velocity $V_e$

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A particle of mass m m m is projected with a velocity v=kVe(k<1) v = k V_e (k < 1) v=kVe​(k<1) from the surface of the Earth.(Ve= \newline (V_e = (Ve​=escape velocity)) \newline )The maximum height above the surface reached by the particle is:

Explanation

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A particle of mass $m$ is projected with a velocity $v = k V_e (k < 1)$ from the surface of the Earth. $\newline (V_e =$ escape velocity $) \newline$ The maximum height above the surface reached by the particle is: