An object is thrown directly away from the surface of the earth with an initial speed $v$. The object…
An object is thrown directly away from the surface of the earth with an initial speed $v$. The object reaches upto a height of $\frac{4}{5} R_E$ from earth's surface, where $R_E$ is radius of the earth. If the escape velocity of the object is $v_E$ then the value of $\frac{v}{v_E}$ is
$4 / 3$
$3 / 4$
$2 / 3$
$4 / 5$
Solution
We know that, maximum height attained by a projectile projected with velocity $v$.
$
h=\frac{v^2 R_E}{2 g R_E-v^2}
$
But given, $h=\frac{4}{5} R_E$
$
\begin{aligned}
& \therefore & \frac{4}{5} R_E & =\frac{v^2 R_E}{2 g R_E-v^2} \\
& \Rightarrow & \frac{4}{5} & =\frac{v^2}{2 g R_E-v^2} \\
& \Rightarrow & 5 v^2 & =8 g R_E-4 v^2 \\
& \Rightarrow & 9 v^2 & =8 g R_E \\
& \Rightarrow & v & =\sqrt{\frac{8}{9} g R_E}=\frac{2}{3} \sqrt{2 g R_E} \\
& \Rightarrow & v & =\frac{2}{3}, v_E \left[\because v_E=\sqrt{2 g R_E}\right]
\\
& \Rightarrow & \frac{v}{v_E} & =\frac{2}{3}
\end{aligned}
$