Rocket's Max Height: Solving for ymax

AI Thread Summary
The rocket accelerates upward at 53.9 m/s² for 7 seconds before running out of fuel. After this period, it continues to ascend until gravity brings it to a stop. To find the maximum height, first calculate the velocity at the end of the fuel burn, then determine how far it ascends after the fuel is exhausted using the formula for motion under constant acceleration. The final height is the sum of the height achieved during propulsion and the height gained during free ascent. The solution involves applying kinematic equations to account for both phases of the rocket's flight.
Turtlie
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Homework Statement


A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 53.9 m/s^2. The acceleration period lasts for time 7.00 s until the fuel is exhausted. After that, the rocket is in free fall.

Find the maximum height ymax reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 m/s^2 .

The Attempt at a Solution


I got 2,641m, but it says that the rocket will still be moving upwards after the fuel is lost. How would I find how far the rocket goes after it runs out of fuel?
 
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From the moment the fuel finishes, only one force acts upon the rocket: gravity.
So find the the velocity the rocket possesses after those 7 seconds of propulsion and then use the constant acceleration formulae to find the height at which the rocket has v=0.


R.
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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