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== Math Problem of the Day ==

If you have a rocket traveling vertically with only the mass of its fuel to carry, with earth sea level gravity, then, how much fuel must you carry to reach a height of 1 kilometer if maintaining a newton of thrust uses up 1 gram of fuel per minute.

Bonus: show the equation with height and fuel per time are both unknowns (variables).

Ok so i made a mistake on this problem. Turns out this was accidentally a trick question. The answer is that any amount of fuel will get you to any height. The reason being that as the fuel is expended the mass of the ship approaches 0, so acceleration approaches infinity. So by the time the rocket is out of fuel it would be moving at the speed of light no matter how much fuel it started with.

As suck I will post a new version of this question and ill make sure i solve it BEFORE I post it.

@SnDer @njha

@snder FYI, QOTO won't respond with the correct answer for 24 hours.

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My Incorrect Attempt 

@QOTO

My solution assumes a constant mass so it's incorrect... How are you actually supposed to do this?

Find the acceleration, turns out to be a constant.
height = 0.5(a - g)t^2
1N = 0.001kg * a
(1N)z = (0.001kg)z * a
a = 1000m/s^2 regardless of fuel

So how long do we need to maintain that accel to get to 1km?
1000 = 0.5(1000-9.81)t^2
t = 1.42120173762 sec (almost sqrt2)
g = .023686696g of fuel

Doesn't seem right because by that logic I can set the fuel consumption to anything and have the same acceleration.

My Incorrect Attempt 

@njha To solve this you need to use a first order differential equation :)

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