Question

Example 28 part (ii)
Example 28. A rocket is set for vertical firing. If the exhaust speed is 1200 m$s^{-1}$, how much gas must be ejected per second to supply the thrust needed (i) to overcome the weight of the rocket, (ii) to give to the rocket an initial vertical upward acceleration of 19.6 m$s^{-2}$? Given mass of rocket = 6000 kg.

Answer 1

Dear student,
Here is the solution of your asked query:
$$\begin{gathered} \mathrm{Upward}\mathrm{thrust}\mathrm{is}\mathrm{given}\mathrm{by}; \\ \mathrm{T}=-\mathrm{u}\frac{\mathrm{dm}}{\mathrm{dt}} \\ \mathrm{where},\mathrm{u}\mathrm{is}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{exhaust}\mathrm{gas}. \\ \frac{\mathrm{dm}}{\mathrm{dt}}\mathrm{is}\mathrm{the}\mathrm{ejected}\mathrm{mass}\mathrm{per}\sec . \\ \left(\mathrm{ii}\right)\mathrm{Let}\mathrm{F}\mathrm{be}\mathrm{the}\mathrm{initial}\mathrm{force}\mathrm{in}\mathrm{upward}\mathrm{direction}. \\ \mathrm{F}=\mathrm{ma}=-\mathrm{u}\frac{\mathrm{dm}}{\mathrm{dt}} \\ \Rightarrow 6000\times 19.6=-1200\times \frac{\mathrm{dm}}{\mathrm{dt}} \\ \Rightarrow \frac{\mathrm{dm}}{\mathrm{dt}}=-98\mathrm{kg}/\sec \end{gathered}$$
Regards