Calculating Maximum Height of a Rocket in Free Fall

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Homework Help Overview

The problem involves calculating the maximum height of a rocket that accelerates upward with a constant acceleration before entering free fall. The scenario includes initial conditions, acceleration values, and the effects of gravity, while ignoring air resistance.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial calculations of distance and velocity during the acceleration phase. Questions arise regarding the subsequent calculations needed to determine the maximum height after the rocket's engines shut off.

Discussion Status

Some participants have provided guidance on how to approach the problem after the acceleration phase, suggesting the use of kinematic equations to find the time to maximum height and the additional distance traveled during free fall. There is ongoing exploration of the calculations and some confusion about the application of different accelerations.

Contextual Notes

Participants are working under the assumption of constant acceleration due to gravity and are attempting to reconcile their calculations with the expected outcomes. There is mention of potential misinterpretations of the equations involved.

Bottomsouth
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Homework Statement



A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 53.9 . The acceleration period lasts for time 5.00 until the fuel is exhausted. After that, the rocket is in free fall.

Find the maximum height reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.8 .


Homework Equations



Distance = 1/2 at^2
Velocity = u + at


The Attempt at a Solution



So far what I get
distance = 1/2 x 53.9 x 5^2 = 673.75 m
V = 0 + 53.9 x 5 = 269.5 m/s

and from there I seem to be stuck. I don't know what else to do, if there is someone to help guide me or of this ditch I'd appreciate it.
 
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That's a good start.

Now how much longer until it gets to max height?

Use the same equation that gives you your speed at main engine shut-off. Only this time it's gravity and not rocket power.

Armed with that time figure then using your distance equation how much further it went. Add the 2 heights together. Done.
 
so I multiplied 53.9 by 9.8 and got 528.22, then added that with 676.75 to get 1201.97. Thats not the answer, I am still off somewhere or mis read?
 
Once its done accelerating.

Solve for t
Vf=V0+at
--------------
Vf=0m/s
V0=269.5m/s
a=-9.8m/s
--------------

Distance formula
d=d0+vo*t+1/2at2
--------------
d0 = 673.75m
v0= 269.5m/s
a=-9.8m/s2
--------------
 
Bottomsouth said:
so I multiplied 53.9 by 9.8 and got 528.22, then added that with 676.75 to get 1201.97. Thats not the answer, I am still off somewhere or mis read?

No. You don't multiply the accelerations together. You use the acceleration of gravity this time instead of the acceleration of the rocket over the 5 seconds.

You figured velocity is 269.5. Divide that by 9.8. That's how many seconds it will continue to rise until velocity is back to 0 - time to max height.

Use that time with y = 1/2*g*t2
to find how much farther it goes against gravity.

That height is what you add to your previous height at main engine cut off to get total height.
 

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