Rocket Stop Time & Distance: 3.156kg, 28.72 m/s

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    Rocket Time
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Discussion Overview

The discussion revolves around calculating the time and distance a rocket, weighing 3.156 kg and moving upwards at 28.72 m/s, takes to come to a stop after its engines are turned off. Participants explore the effects of gravity and air resistance on the rocket's motion, focusing on kinematic equations and the implications of neglecting drag.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the time to stop using the equation t = (0 - 28.72) / -9.8, resulting in approximately 2.9 seconds.
  • Another participant seeks to determine the distance traveled during the 2.9 seconds, proposing a formula d = v^2 / (2ug) and calculating it to be 42.08 meters.
  • There is a discussion about the kinematic equation V^2 = U^2 + 2gs, with participants confirming the use of initial velocity and gravitational acceleration.
  • Some participants suggest using average speed to find the distance, given the constant acceleration.
  • Concerns are raised about neglecting air resistance (drag), with one participant arguing that drag is a complex factor that may not significantly affect the calculations for this scenario.
  • Another participant expresses relief at not needing to account for drag in their current calculations, indicating a willingness to learn about it later.

Areas of Agreement / Disagreement

Participants generally agree on the basic kinematic calculations for time and distance but disagree on the relevance and complexity of including drag in the analysis. Some argue it should be considered, while others believe it complicates the situation unnecessarily.

Contextual Notes

Participants express varying levels of understanding regarding kinematics and the implications of drag, indicating that some assumptions about the simplicity of the problem may not hold when considering real-world factors.

Max CR
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A rocket is moving upwards at 28.72 meters per second. Then, its engines are turned off. How much time will it take for it to come to a stop if it is traveling upwards exactly straight? Also, how far would the rocket have had traveled after its engines were shut off? The weight of the rocket is 3.156 kilograms.

Thanks. This is a personal project that I am working on and is not a homework question.
 
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Ignoring air resistance it's quite easy - including air resistance is a little trickier but possible if you don't mind some programming or a spreadsheet.

Any falling object accelerates at the same rate 9.8m/s^2 and any rising object (once the motor has stopped) slows by 9.8m/s^2

Then it's simply v=u+at since v=0 when it stops you simply have t = 0-28.72/-9.8 = 2.9s
 


Ok great. That told me how long it will take. Now i just need to know how far it will travel in those 2.9 seconds. Thanks
 


Any ideas?
 


Ok. I believe i found a formula but I am not sure. The formula is

d = v^2/2ug = (28.72 m/s)^2/2(9.8 m/s) = 42.08 meters

Is this correct?
 


Yes the equation is normally written as, V^2=U^2 + 2 g s
Again v=0 and u=28.72m/s g=9.8m/s/s
There is a sticky in the intro physics with all these equations
 


Ok. I have the following

0 = (28.72 m/s)^2 + 2(9.8 m/s)(2.9 seconds)

Now how do I find how far the rocket traveled? I am trying to figuer out how far the rocket will travel in the 2.9 seconds that it takes for it to stop and come back to the ground.

Thanks
 


Sorry 's' is displacement ie. distance (for slightly complicated reasons)
0 = (28.72 m/s)^2 + 2(9.8 m/s/s) * distance
so rearranged
distance = 0-(28.72 m/s)^2 / 2 * (9.8 m/s/s) = 42.08m

This is the distance up (ie until speed is zero) the total time and total distance is twice this.
 


You can also take the average speed, since the acceleration is constant: (28.72)/2*2.9
 
  • #10


Perfect! Thank you!
 
  • #11


OK. But you can't neglect drag.
 
  • #12


Ok now how do I calculate drag?
 
  • #13


Phrak said:
OK. But you can't neglect drag.


The guy doesn't even know basic kinematics and you're suggesting he calculate drag?

Don't worry about drag, it's a much, much more complicated creature and in most cases makes little to no difference.
 
  • #14


Oh good. I am glad to hear that I don't have to redo all of the calculations regarding the rockets altitude. Haha. Thanks.

I will learn how to calculate drag later in my college years. Thanks though.
 
  • #15


maverick_starstrider said:
The guy doesn't even know basic kinematics and you're suggesting he calculate drag?

Don't worry about drag, it's a much, much more complicated creature and in most cases makes little to no difference.

It's a rocket not a rock.


...and why would you think I'm asked the OP to answer his own question?
 
Last edited:

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