anonymous
  • anonymous
A model rocket is launched vertically with an engine that is ignited at time t = 0. The engine provides an upward acceleration of 30 m/^2 for 2 seconds. Upon reaching its maximum height, the rocket deploys a parachute, and then descends vertically toward the ground. A) Determine the speed of the rocket after the 2s firing of the engine B) What maximum height will the rocket reach? C) At what time after t=0 will the maximum height be reached?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
i got this but i'm not sure: A.) i assume that the velocity is equal to the speed: a=v/t v=at =(30 m/s^2)(2s) =60 m/s
anonymous
  • anonymous
lol, i'm wrong??? sorry @Katie5

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theEric
  • theEric
No, that link starts out wrong!!
theEric
  • theEric
@Data_LG2 did part A correctly!
anonymous
  • anonymous
really???????
theEric
  • theEric
Yes, haha! I don't know why the responder in the other link decided differently.
anonymous
  • anonymous
lol, can you continue B, I'm not sure how to do that:p
theEric
  • theEric
I would say that the problem means that the rocket has an acceleration of \(30\ [m/s^2]\). If it meant that it has an acceleration of \(30\ [m/s^2]\) in addition to the acceleration of gravity, then the total acceleration would be less than \(30\ [m/s^2]\). And air resistance would lower that even more. So I don't think that \(40\ [m/s]\) is an option at all.
theEric
  • theEric
For part B, I would use \(d=d_0+v_0\ \Delta t+\frac{1}{2}a\ (\Delta t)^2\). @Katie5 , does that look familiar?
anonymous
  • anonymous
yes, it's on one of the equations from our review sheet
theEric
  • theEric
Cool! Do you know what the variables are, then?
anonymous
  • anonymous
yupp! i am just struggling to find the answer. my teacher said we need to use the quadratic equation twice to find the answer and i got really lost
theEric
  • theEric
Quadratic equation? Which one is that? Is it one of the physics equations? Was it the one I just posted?
theEric
  • theEric
I can see you having to use that now and later for another part. That sounds better than I thought at first.
anonymous
  • anonymous
okay so i do use it?
theEric
  • theEric
Well, you need to know what the variables are! If there is a subscript of \(0\), like \(d_0\), then it means we are talking about that value at the beginning of the interval. If there is no subscript \(0\), then it means the value is for at the end of the interval. \(d\equiv\) distance \(v\equiv\) velocity \(a\equiv\) acceleration So \(d\) is the final distance, and \(d_0\) is the starting distance. We'll just say that the distance is measured from the ground. So, since the rocket starts there, \(d_0=0\). Make sense?
theEric
  • theEric
Do you see what you need to do? Just substitute in the values appropriately. \(v_0\): what is the initial velocity? \(a\): what is the acceleration? \(\Delta t\): what is the time interval? Then you can solve for \(d\), the final distance - when the rocket stops accelerating. You find that, and then you need to find how much farther it goes! I see what your teacher means - we'll be using that same equation again. I'll show you why if you'd like, but first you must get what I wrote above!
anonymous
  • anonymous
ok, i think i understand. thank you!
theEric
  • theEric
Cool! That's good! In the next interval the rocket is continuing upwards, but slowing down due to gravitational acceleration. At the beginning, the rocket is already up in the air, so you know \(x_0\neq0\). In fact, \(x_0>0\). It is the \(x\) value from the last time you used the equation. Here is a not-to-scale diagram: |dw:1377877119716:dw| If you wanted to keep your \(x\)'s and other variables more clear, you could label the ends of the intervals and use those labels in subscripts.
theEric
  • theEric
|dw:1377877543055:dw| Going by that (only if you want - I think it is easier to keep track of things), then you already solved for \(x_1\) where \(x_1=x_0+v_0\ \Delta t_{0\rightarrow 1}+\frac{1}{2}a_\text{thrusting}\ \Delta t_{0\rightarrow 1}^2\). Now I'll give you another equation: \(v^2=v_0^2+2\ a\ \Delta d=v_0^2+2\ a\ (d-d_0)\). Solving for \(d\), that means that \(v^2=v_0^2+2\ a\ (d-d_0)\\\implies \dfrac{v^2-v_0^2}{2\ a}+d_0=d\) Regular \(d\) is at the end of the interval, \(d_0\) was at the beginning of the interval. Still going by those labels, the top distance is: \(\dfrac{v_2^2-v_1^2}{2\ a_\text{gravity only}}+d_1=d_2\) \(v_2=0\), because the rocket has no velocity at the tope (it's between going up and going down).
theEric
  • theEric
Continuing to use that notation, you can find the answer to part c: total time going up. It took two seconds to go up while the rocket was in thrust. Then you can use the definition of acceleration to find the time for the rest of it. Now, to pick the formulas, I'm going through my head and looking at what formulas have the variables I have and need, and no variables I don't have. You can do the same with the equations you were given! :) Sometimes there is a variable you don't know but that you can calculate. Then you calculate that variable and you can use it in a formula!

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