If a particle is fired vertically upwards from the surface of the earth and reaches a height of 6400km, the initial velocity of the particle is : (Assume R=6400km and 2 g '10 ms^-2)

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4 km/sec

2 km/sec

8 km/sec

16 km/sec

answer is C.

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Detailed Solution

According to law of conservation of energy,
$\begin{aligned} \frac{1}{2} {mv}^{2} = \frac{mgh}{1 + \frac{h}{R}} ∴ v^{2} = \frac{2 gh}{1 + \frac{h}{R}} = \frac{2 \times 10 \times 6 .4 \times 10^{6}}{1 + \frac{R}{R}} \\ = \frac{2 \times 10 \times 6 .4 \times 10^{6}}{2} ∴ v = \sqrt{64 \times 10^{6}} = 8 km / \sec . \end{aligned}$