How Do You Calculate Rocket Acceleration with Limited Information?

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To calculate the rocket's acceleration during the first 16 seconds, consider that the rocket experiences constant acceleration while the motor is active and then decelerates under gravity after the motor stops. The initial velocity is zero, and the displacement during the first phase can be expressed as h1 = (1/2) * a * t1^2, where "a" is the unknown acceleration. After the motor stops, the rocket continues to rise for an additional 4 seconds, with its displacement calculated as h2 = v1 * t2 - (1/2) * g * t2^2, where v1 is the velocity at the end of the motor phase. By summing the displacements from both phases and setting them equal to the total altitude of 5100 m, a single equation can be formed to solve for the acceleration "a."
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A 1000kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance.


What's the rocket's acceleration during the first 16 seconds?



I posted this in another thread but I guess nobody is looking at it... if posting this is against the rules then I guess ban me/lock thread but PLEASE how do I even start? Initial velocity is zero, I know time, but with no displacement (the 5100 is for the very end after the motor stopped) or final velocity (when the motor stops) I just can't get it. I used all 3 equations and the best I can do is get acceleration in terms of final postion (the one where the motor stops).

:cry:
 
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The 1000 kgs seems to suggest some force analysis.
 
It's from a chapter before forces are introduced so it shouldn't have anything with the problem...
 
FancyNut said:
A 1000kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance.


What's the rocket's acceleration during the first 16 seconds?



Initial velocity is zero, I know time, but with no displacement (the 5100 is for the very end after the motor stopped) or final velocity (when the motor stops) I just can't get it. I used all 3 equations and the best I can do is get acceleration in terms of final postion (the one where the motor stops).

:cry:

Try to think logically instead of plugging in data into formulas... What happened to the rocket? It rose with constant acceleration for t1=16 s, after that the motor stopped but gravity still acted on it, so it rose with deceleration of g for t2=4 s.
You do not know the acceleration during the first t1 time, so it is "a".
The velocity rose to v1=a*t1, and the displacement is h1=a/2*t1^2.
Now the second period comes, for t2=4 seconds, with acceleration -g and "starting" velocity vo=v1. The displacement is h2=vo*t2 -g/2 *t2^2= a*t1*t2-g/2*t2^2.
The sum of both displacements is h = h1+h2 =5100 m. You have one equation with one unknown, it is the acceleration, a.

ehild
 
ehild said:
Try to think logically instead of plugging in data into formulas... What happened to the rocket? It rose with constant acceleration for t1=16 s, after that the motor stopped but gravity still acted on it, so it rose with deceleration of g for t2=4 s.
You do not know the acceleration during the first t1 time, so it is "a".
The velocity rose to v1=a*t1, and the displacement is h1=a/2*t1^2.
Now the second period comes, for t2=4 seconds, with acceleration -g and "starting" velocity vo=v1. The displacement is h2=vo*t2 -g/2 *t2^2= a*t1*t2-g/2*t2^2.
The sum of both displacements is h = h1+h2 =5100 m. You have one equation with one unknown, it is the acceleration, a.

ehild


*gives ehild a box of cookies*

:D :D :D

I guess next time I shouldn't be so lazy and just expect to plug in numbers in a given formula... :redface:
 
FancyNut said:
*gives ehild a box of cookies*

:D :D :D

I guess next time I shouldn't be so lazy and just expect to plug in numbers in a given formula... :redface:


Great! and thanks for the virtual cookies. They were yummy :smile:

ehild
 
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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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