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

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(A) By the law of conservation of energy,the total energy at the surface of the earth is equal to the total energy at the maximum height $h$.At the su

Vedclass · Accepted Answer

$\frac{u^2 R}{2 g R - u^2}$

Vedclass · Answer

(A) By the law of conservation of energy,the total energy at the surface of the earth is equal to the total energy at the maximum height $h$.At the surface: $E_i = -\frac{GMm}{R} + \frac{1}{2}mu^2$At maximum height $h$: $E_f = -\frac{GMm}{R+h} + 0$Equating $E_i = E_f$: $-\frac{GMm}{R} + \frac{1}{2}mu^2 = -\frac{GMm}{R+h}$Dividing by $m$ and rearranging: $\frac{GM}{R+h} = \frac{GM}{R} - \frac{u^2}{2}$Using $GM = gR^2$: $\frac{gR^2}{R+h} = gR - \frac{u^2}{2} = \frac{2gR - u^2}{2}$$\frac{R+h}{R^2} = \frac{2g}{2gR - u^2}$$R+h = \frac{2gR^2}{2gR - u^2}$$h = \frac{2gR^2}{2gR - u^2} - R = \frac{2gR^2 - 2gR^2 + u^2R}{2gR - u^2} = \frac{u^2R}{2gR - u^2}$

$A$ body is thrown from the surface of the earth with velocity $u \ m \ s^{-1}$. The maximum height in meters above the surface of the earth up to which it will reach is ($R=$ radius of the earth,$g=$ acceleration due to gravity).

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