Rocket Fuel Consumption: Calculation of 80kN Takeoff Rate

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icymage
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The weight of a rocket is 80kN at the instant it takes off , and the products of the combusion are ejected at a velocity of 0.60 kms-1 relative to the rocket. Calculate the rate at which the rocket must be consuming fuel in order to take off vertically. Ans: 134kgs-1

Cant find a solution despite numerous attempt , anyone can help ?
 
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hmm i tried using the equation f = m(dv/dt) + v (dm/dt) and p = mv and i can't get the answer. Anw this isn't coursework/schoolwork it is a question i stumbled upon on a old textbook of mine.
 
icymage said:
hmm i tried using the equation f = m(dv/dt) + v (dm/dt) and p = mv and i can't get the answer. Anw this isn't coursework/schoolwork it is a question i stumbled upon on a old textbook of mine.
Take a look at this excellent https://www.physicsforums.com/showthread.php?t=199087" by D_H, and see if it helps clarify things.
 
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Welcome to PF!

icymage said:
hmm i tried using the equation f = m(dv/dt) + v (dm/dt) and p = mv and i can't get the answer.

Hi icymage! Welcome to PF! :smile:

This isn't really a caluclus question … it would be if you were asked to find the fuel used to reach a certain time, for example … but you're only asked for the initial fuel used while the speed is zero.

Hint: how much acceleration does the fuel have to produce before the rocket lifts off?

how much mass and velocity produce that acceleration? :smile: