A 5000 kg rocket is set for vertical firing. The exhaust speed is$\ {800\ ms}^{- 1}$. To give an initial upward acceleration of ${20\ ms}^{- 2}$, the amount of gas ejected per second to supply the needed thrust will be $(g = {10\ ms}^{- 2})$

Question

Dr. Abubacker · Accepted Answer

(b) Thrust force by rocket
$F_{t} = v_{r}\left( - \frac{dm}{dt} ight)\ (upwards)$
Weight of the rocket
$w = mg\ (downwards)$
Net force on the rocket
$F_{net} = f_{t} - w$
$\Longrightarrow \ \ \ \ ma = v_{r}\left( \frac{- dm}{dt} ight) - mg$
$\Longrightarrow \ \ \ \ \ ma = v_{r}\left( \frac{- dm}{dt} ight) = \frac{m(g + a)}{v_{r}}$
$	herefore\ \ $Rate of the ejected per second
$= \frac{5000(10 + 20)}{800} = \frac{5000 	imes 30}{800}$
$= 187.5\ {kgs}^{- 1}$

Q. A 5000 kg rocket is set for vertical firing. The exhaust speed is$\ {800\ ms}^{- 1}$. To give an initial upward acceleration of ${20\ ms}^{- 2}$, the amount of gas ejected per second to supply the needed thrust will be $(g = {10\ ms}^{- 2})$

Explanation

Reviewed by Dr. Abubacker

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