A body is thrown from the surface of the eart | vedclass

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(B) Using the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.Initial energy at surfac

Vedclass · Accepted Answer

$\frac{V^{2}R}{2gR-V^{2}}$

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(B) Using the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.Initial energy at surface: $E_i = \frac{1}{2}mV^2 - \frac{GMm}{R}$Final energy at height $h$: $E_f = 0 - \frac{GMm}{R+h}$Equating $E_i = E_f$:$\frac{1}{2}mV^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$$\frac{1}{2}V^2 = GM \left( \frac{1}{R} - \frac{1}{R+h} \right)$Since $GM = gR^2$:$\frac{1}{2}V^2 = gR^2 \left( \frac{R+h-R}{R(R+h)} \right) = gR \left( \frac{h}{R+h} \right)$$\frac{V^2}{2gR} = \frac{h}{R+h}$$V^2(R+h) = 2gRh$$V^2R + V^2h = 2gRh$$V^2R = h(2gR - V^2)$$h = \frac{V^2R}{2gR - V^2}$

$A$ body is thrown from the surface of the earth with velocity $V \ m/s$. The maximum height above the earth's surface up to which it will reach is ($R =$ radius of earth,$g =$ acceleration due to gravity).

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