A rocket is launched vertical from the surface of the earth of radius R with an initial speed v. If atmospheric resistance is neglected. the maximum height attained the rocket is

h = \frac{R}{(\frac{2 g R}{v^{2}} - 1)}

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h = \frac{R}{(\frac{2 g R}{v^{2}} + 1)}

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h = \frac{R^{2}}{(\frac{2 g R}{v^{2}} - 1)}

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h = \frac{R^{2}}{(\frac{2 g R}{v^{2}} + 1)}

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Solution

The correct option is A $h = \frac{R}{(\frac{2 g R}{v^{2}} - 1)}$
KE of launched rocket = change in PE of rocket on reaching max height
$\frac{1}{2} M V^{2} = M g h [\frac{R}{R + h}]$
$\frac{R + h}{R} = \frac{2 g R}{V^{2}}$
$\frac{R}{h} = \frac{2 g R}{V^{2}} - 1$
$h = \frac{R}{(\frac{2 g R}{V^{2}} - 1)}$

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A rocket is launched vertical from the surface of the earth of radius R with an initial speed v. If atmospheric resistance is neglected. the maximum height attained the rocket is

The correct option is A h=R(2gRv2−1)KE of launched rocket = change in PE of rocket on reaching max height12MV2=Mgh[RR+h]R+hR=2gRV2Rh=2gRV2−1h=R(2gRV2−1)