Question

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

• A. $127.5$kg s${}^{-1}$
• B. $137.5$kg s${}^{-1}$
• C. $187.5$kg s${}^{-1}$
• D. $185.5$kg s${}^{-1}$

Answer 1

Answer: (C)

$187.5$kg s${}^{-1}$

We know that $F=\frac{\delta p}{\delta t}$
$\delta p=$rate of ejection$\times \delta t\times v$
$F=$rate of ejection $\times v$.
Also , $F-mg=ma\Rightarrow F=m(g+a)$
Therefore, rate of ejection $=\frac{m(g+a)}{v}=\frac{5000\times 30}{800}=187.5$kg s${}^{-1}$.