Question

Find the least rate which a rocket must consume fuel to be able to take off vertically. Given: Mass of rocket at the instant of takeoff is \[3\] tons and ejection velocity of products of combustion is \[600m/s\] . g\[ = 9.8m/{s^2}\].

Answer 1

Hint: To solve this question we have to know the formula of velocity of products of combustion. And also the rate of consumed fuel. Here m is the mass of the rocket and a is equal to the acceleration of the rocket. Here g is the gravitation. And the formula is \[F = ma\] . Now we have to convert the mass value of the rocket from tons to cubic meters.Otherwise we can not solve the problem.

Complete step by step answer:
We know that \[F = ma\] ,
We also know that,
\[mg = V\dfrac{{dm}}{{dt}}\]
Where g is equal to the gravitation. Velocity is equal to V and m is equal to the mass of the rocket at the instant of take-off .
\[\dfrac{{dm}}{{dt}} = \dfrac{{mg}}{V} \\
\Rightarrow\dfrac{{dm}}{{dt}} = \dfrac{{3 \times {{10}^3} \times 9.8}}{{600}} \\
\Rightarrow\dfrac{{dm}}{{dt}} = 49kg/s\]
Here, \[3tons = 3 \times {10^3}{m^3}\] \[\left( {} \right.\] we know \[\left. {} \right)\]

The least rate of a rocket must consume fuel to be taken off vertically is \[49kg/s\].

Note: We always forget to convert the value of mass which is given in the question in tons. We have to convert that into cubic meters. We have to calculate this in one same unit whether it can be an S.I unit or in a C.G.S unit. Otherwise the calculation will become wrong and the mark will also be deducted for this mistake. Not only this tons unit but also whenever there will be the mixture of units always we have to convert those and make them all in similar units.