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A body is projected up with a velocity equal to $\frac{3}{4}$ th of the escape velocity from the surface of the earth. The height it reaches is (Radius of the earth $=R$ )
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Verified Answer
The correct answer is:
$\frac{9 R}{7}$
Velocity of projection $v=\frac{3}{4} v_e$
where, $v_e=$ escape velocity $=\sqrt{2 g R}$
Maximum height attained by the body
$$
\begin{aligned}
h & =\frac{v^2 R}{2 g R-v^2} \\
& =\frac{\left(\frac{3}{4} \sqrt{2 g R}\right)^2 \times R}{2 g R-\left(\frac{3}{4} \sqrt{2 g R}\right)^2}=\frac{\frac{9}{16} \times 2 g R \times R}{2 g R-\frac{9}{16} \times 2 g R} \\
& =\frac{\frac{9}{16} \times 2 g R^2}{2 g R\left(1-\frac{9}{16}\right)}=\frac{\frac{9}{16} R}{\frac{7}{16}} \\
& =\frac{9}{7} R
\end{aligned}
$$
where, $v_e=$ escape velocity $=\sqrt{2 g R}$
Maximum height attained by the body
$$
\begin{aligned}
h & =\frac{v^2 R}{2 g R-v^2} \\
& =\frac{\left(\frac{3}{4} \sqrt{2 g R}\right)^2 \times R}{2 g R-\left(\frac{3}{4} \sqrt{2 g R}\right)^2}=\frac{\frac{9}{16} \times 2 g R \times R}{2 g R-\frac{9}{16} \times 2 g R} \\
& =\frac{\frac{9}{16} \times 2 g R^2}{2 g R\left(1-\frac{9}{16}\right)}=\frac{\frac{9}{16} R}{\frac{7}{16}} \\
& =\frac{9}{7} R
\end{aligned}
$$
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