Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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$= 9.8m/{s^2}$.

Last updated date: 20th Jul 2024
Total views: 382.2k
Views today: 6.82k
Verified
382.2k+ views
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.

We know that $F = ma$ ,
$mg = V\dfrac{{dm}}{{dt}}$
$\dfrac{{dm}}{{dt}} = \dfrac{{mg}}{V} \\ \Rightarrow\dfrac{{dm}}{{dt}} = \dfrac{{3 \times {{10}^3} \times 9.8}}{{600}} \\ \Rightarrow\dfrac{{dm}}{{dt}} = 49kg/s$
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 $49kg/s$.