Question

16. Find the minimum rate at which fuel must be consumed by a rocket so as to be able
to take off vertically. Given that the total mass at the time of take off is 3000 kg and
the velocity of the fuel expelled is 300 m/s. (g = 9.8 m/s]

(I challenge you to do this )​

Answer 1

Answer:

98 kg / s

Explanation:

Thus total mass = 3000 kg

Exhaust speed of rocket v     = 300 m/s  

Force = mass×gravity = 3000×9.8m/s= 29400N

Also we know force of an object = mass×  v / t  as initial velocity = 0

So F=v  ​  × dm / dt where    dm / dt   is the rate of fuel burn.

dm / dt​    =  F / v = 29400 / 300

                 = 98 kg / s