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Q.

A rocket is fired vertically with a speed half of escape speed from the earth’s surface. How far from the earth surface does the rocket go before returning to the earth? (Radius of earth is R)

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a

R/3

b

R/2

c

R

d

2R

answer is D.

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Detailed Solution

$\frac{1}{2} m {(\frac{V_{E}}{2})}^{2} = \frac{m g h}{1 + \frac{h}{R}}$

$\frac{1}{2} (\frac{2 g R}{4}) = \frac{g h}{(1 + \frac{h}{R})} \Rightarrow 4 h = R + h \Rightarrow h = \frac{R}{3}$

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