On the pole of the Earth a body is imparted velocity $v_{0}$ directed vertically up. Knowing the radius of the Earth and the free-fall acceleration on its surface, the height to which the body will ascend is given as $h = \frac{R v_{0}^{2}}{x g R - v_{0}^{2}}$ . The air drag is to be neglected. Find $x$ .

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Solution

Let the mass of the body be $m$ and height it reaches is $h$ .
From energy Conservation,
$- \frac{G M m}{R} + \frac{1}{2} m v_{o}^{2} = - \frac{G M m}{R + h} + 0$
$\frac{G M}{R + h} = \frac{2 G M - v_{o}^{2} R}{2 R}$
$h = \frac{R v_{o}^{2}}{2 g R - v_{o}^{2}}$
So, $x = 2$

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On the pole of the Earth a body is imparted velocity v0 directed vertically up. Knowing the radius of the Earth and the free-fall acceleration on its surface, the height to which the body will ascend is given as h=Rv20xgR−v20. The air drag is to be neglected. Find x.

Let the mass of the body be m and height it reaches is h.From energy Conservation,−GMmR+12mv2o=−GMmR+h+0GMR+h=2GM−v2oR2Rh=Rv2o2gR−v2oSo, x=2

Let the mass of the body be $m$ and height it reaches is $h$ .
From energy Conservation,
$- \frac{G M m}{R} + \frac{1}{2} m v_{o}^{2} = - \frac{G M m}{R + h} + 0$
$\frac{G M}{R + h} = \frac{2 G M - v_{o}^{2} R}{2 R}$
$h = \frac{R v_{o}^{2}}{2 g R - v_{o}^{2}}$
So, $x = 2$