The escape velocity of a body on the earth's surface is $v_e$.A body is thrown up with a speed $\sqrt{5}…
The escape velocity of a body on the earth's surface is $v_e$.A body is thrown up with a speed $\sqrt{5} v_e$.Assuming that the sun and planets do not influence the motion of the body, velocity of the body at infinite distance, is :
$0$
$v_e$
$\sqrt{2} v_e$
$2v_e$
Solution
Since velocity of projection $(v)$ is greater than the escape velocity $\left(v_e\right)$, therefore at infinite distance the body moves with a velocity
$
\begin{aligned}
v^{\prime} & =\sqrt{v^2-v_e^2} \\
\therefore \quad v^{\prime} & =\sqrt{\left(\sqrt{5} v_e\right)^2-v_e^2}=2 v_e
\end{aligned}
$