If a body is projected vertically from the su | vedclass

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(C) Using the law of conservation of energy, the total energy at the surface equals the total energy at the maximum height $h$.At the surface: $E_i =

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$6400$

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(C) Using the law of conservation of energy, the total energy at the surface equals the total energy at the maximum height $h$.At the surface: $E_i = K_i + U_i = \frac{1}{2}mv^2 - \frac{GMm}{R}$At maximum height: $E_f = K_f + U_f = 0 - \frac{GMm}{R+h}$Equating $E_i = E_f$: $\frac{1}{2}mv^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$Dividing by $m$ and using $GM = gR^2$: $\frac{v^2}{2} - gR = - \frac{gR^2}{R+h}$Rearranging: $\frac{gR^2}{R+h} = gR - \frac{v^2}{2} = gR(1 - \frac{v^2}{2gR})$$\frac{R}{R+h} = 1 - \frac{v^2}{2gR} = 1 - \frac{(8000)^2}{2 	imes 10 	imes 6400 	imes 10^3} = 1 - \frac{64 	imes 10^6}{128 	imes 10^6} = 1 - 0.5 = 0.5$$\frac{R}{R+h} = 0.5 \implies R+h = 2R \implies h = R$Since $R = 6400 \,km$, the maximum height $h = 6400 \,km$.

If a body is projected vertically from the surface of the earth with a speed of $8000 \,ms^{-1}$, then the maximum height reached by the body is (Radius of the earth $= 6400 \,km$ and acceleration due to gravity $= 10 \,ms^{-2}$) (in $km$)

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