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Standard XII

Physics

v-t Graph

A projectile ...

Question

A projectile is fired upward from the surface of the earth with a velocity $k v_{e}$ , where $v_{e}$ is escape velocity and $k < 1$ . If $R$ is the radius of earth, then neglecting the air resistance, the maximum height to which it will rise is

\frac{k}{R^{2} - k^{2}}

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\frac{R}{R^{2} {- k}^{2}}

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\frac{k}{1 - R^{2}}

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\frac{k^{2} R}{1 - k^{2}}

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Solution

The correct option is D $\frac{k^{2} R}{1 - k^{2}}$
If a body is projected from the surface of earth with a velocity $v$ and reaches a height $h$ , then by conservation of energy (relative to surface of earth)
$\frac{1}{2} m v^{2} = \frac{m g h}{[1 + (\frac{h}{R})]}$
In this case, $v = k v_{e} = k \sqrt{2 g R}$
So, $\frac{1}{2} k^{2} (2 g R) = \frac{g h}{[1 + (\frac{h}{R})]} o r k^{2} = \frac{h}{R + h}$
$o r k^{2} R + k^{2} h = h \Rightarrow h = \frac{k^{2} R}{1 - k^{2}}$

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