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A body is projected vertically upwards from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth, maximum height attained by the body from the surface of the earth is
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Verified Answer
The correct answer is:
$\frac{R}{3}$
Here, maximum height attained by a projectile
$$
h=\frac{v^{2} R}{2 g R-v^{2}}...(i)
$$
Velocity of body $=$ half the escape velocity
i.e., $v=\frac{v_{e}}{2}$
or $v=\frac{\sqrt{2 g R}}{2} \Rightarrow v^{2}=\frac{2 g R}{4} \Rightarrow v^{2}=\left(\frac{g R}{2}\right)$
Now, putting value of $v^{2}$ in Eq. (i), we get
Height, $h=\frac{\frac{g R}{2} \cdot R}{2 g R-\frac{g R}{2}}=\frac{\frac{g R^{2}}{2}}{\frac{3 g R}{2}}=\frac{R}{3}$
$$
h=\frac{v^{2} R}{2 g R-v^{2}}...(i)
$$
Velocity of body $=$ half the escape velocity
i.e., $v=\frac{v_{e}}{2}$
or $v=\frac{\sqrt{2 g R}}{2} \Rightarrow v^{2}=\frac{2 g R}{4} \Rightarrow v^{2}=\left(\frac{g R}{2}\right)$
Now, putting value of $v^{2}$ in Eq. (i), we get
Height, $h=\frac{\frac{g R}{2} \cdot R}{2 g R-\frac{g R}{2}}=\frac{\frac{g R^{2}}{2}}{\frac{3 g R}{2}}=\frac{R}{3}$
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