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A rocket set for vertical firing weighs 50 kg and contains 450 kg of fuel. The product of combustion can have a maximum velocity of 2 km/s. What should be the minimum rate of fuel consumption to give it an acceleration of $20 m / s^{2}$ ? Take $g = 9.8 m / s^{2} .$

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4.74 kg/s

6.25 kg/s

3.45 kg/s

7.45 kg/s

answer is D.

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Detailed Solution

$ma = - mg + V_{r} (- \frac{dm}{dt}) ∴ - \frac{dm}{dt} = \frac{m (a + g)}{V_{r}} = \frac{(450 + 50) (20 + 9 .8)}{2 \times 10^{3}} kg / s \Rightarrow - \frac{dm}{dt} = 7 .45 kg / s$

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