A rocket is launched vertically from the surface of earth with an initial velocity v. How far above the surface of earth it will go? Neglect the air resistance.

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

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

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

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

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Solution

The correct option is D $R {(\frac{2 g R}{v^{2}} - 1)}^{- 1}$
By conservation of energy,
Kinetic energy of launched rocket $=$ change in potential energy of rocket at maximum height.
$\to$ Ig $h$ is maximum height reached:
$\frac{1}{2} m v^{2} = m g h (\frac{R}{R + h}) \Rightarrow \frac{R + h}{h} = \frac{2 g R}{v^{2}} \Rightarrow h = R (\frac{2 g R}{v^{2}} - 1)^{- 1}$

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A rocket is launched vertically from the surface of earth with an initial velocity v. How far above the surface of earth it will go? Neglect the air resistance.

The correct option is D R(2gRv2−1)−1By conservation of energy,Kinetic energy of launched rocket = change in potential energy of rocket at maximum height.→ Ig h is maximum height reached:12mv2=mgh(RR+h)⇒R+hh=2gRv2⇒h=R(2gRv2−1)−1