A body is projected vertically upwards from the surface of a planet of radius \(R\) with a velocity equal to…
A body is projected vertically upwards from the surface of a planet of radius \(R\) with a velocity equal to half the escape velocity of that planet. Then, the maximum height attained by the body is
\(\frac{R}{3}\)
\(\frac{R}{2}\)
\(\frac{R}{4}\)
\(\frac{R}{5}\)
Solution
If a body is projected vertically upwards with a velocity of \(v\) from the surface of planet, then maximum height attained by the body is
\(h=\frac{v^2 R}{2 g R-v^2}\) ...(i)
But given that,
\(\begin{aligned}
v & =\frac{v_e}{2}=\frac{\sqrt{2 g R}}{2} \Rightarrow v^2=\frac{2 g R}{4} \\
v^2 & =\frac{g R}{2} \quad \ldots (ii)
\end{aligned}\)
From Eqs. (i) and (ii), we get
\(h=\frac{\frac{g R}{2} \cdot R}{2 g R-\frac{g R}{2}}=\frac{\frac{g R^2}{2}}{\frac{3 g R}{2}} \Rightarrow h=\frac{R}{3}\)