$v_0 \sin \theta_0 \left( \frac{v_0 \sin \theta_0}{g} \right) - \frac{1}{2}g \left( \frac{v_0 \sin \theta_0}{g} \right)^2$
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Step 1: The given expression is: $v_0 \sin \theta_0 \left( \frac{v_0 \sin \theta_0}{g} \right) - \frac{1}{2}g \left( \frac{v_0 \sin \theta_0}{g} \right)^2$ Show more…
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Let the mass of the body be $m$ and let it go upto a height $h$. From conservation of mechanical energy of the system $$ -\frac{\gamma M m}{R}+\frac{1}{2} m v_{0}^{2}=\frac{-\gamma M m}{(R+h)}+0 $$ Using $\frac{\gamma M}{R^{2}}=g$, in above equation and on solving we get, $$ h=\frac{R v_{0}^{2}}{2 g R-v_{0}^{2}} $$
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