A model rocket

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Hello loves, I know there are post very very similar to this, but I just needed some help understanding it. any help is truly appreciated.

1. Homework Statement

A model rocket is launched straight upward with an initial speed of 50 m/s. It accelerates with a constant upward acceleration of 30 m/s2 until its engines stop at an altitude of 150 m. How long is the rocket in the air?

Vi= 50m/s
Vf= ??
a= 30 m/s^2
Change in y= 150

Homework Equations


upload_2015-1-22_23-33-37.png


The Attempt at a Solution


Alright so, I think to solve this problem, its a 2 step thing, this is caused because of the 2 accelerations, one from the rocket's fuel, and then once it empties, the second one from earth's gravitational acceleration. I know I have 3 givens so I can find the other 2 variables. Alright this has led me to choosing the fourth and first equation. Although im not sure if i can replace the delta x with delta Y, because the rocket would be going up into the air vertically, so it would be y instead of x right? this is where i am stuck, i dont really know what to do once i get to this point.

a preemptive thank you guys for the help
 

Answers and Replies

  • #2
gneill
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The particular symbols used to represent quantities aren't as important as the equation matching the given situation. Here you have motion in a straight line with constant acceleration(s). The equations you've shown match the scenario, so feel free to change Δx to Δy, or y, or z for that matter.
 
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  • #3
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Alright, so, the Δx is replaceable, that is a good thing to know. Alright so I have been manipulating all of these equations to try to isolate time. So for the first part, I am trying to find out how long it takes to get to 150 meters. I have yet to find this equation, or been able to manipulate these 4 to give it to me. for the second part, I can find the Final velocity of the first part, and make that the initial velocity, and the final velocity of the second one will be 0 correct? because it would hit the floor coming to a stop. So once again i have to isolate t, but this time i have the following variables:
Yi = 150 meters
Yf= 0
a=-9.8 m/s^2
Vi= Vf of part 1
 
  • #4
haruspex
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Alright, so, the Δx is replaceable, that is a good thing to know. Alright so I have been manipulating all of these equations to try to isolate time. So for the first part, I am trying to find out how long it takes to get to 150 meters. I have yet to find this equation, or been able to manipulate these 4 to give it to me. for the second part, I can find the Final velocity of the first part, and make that the initial velocity, and the final velocity of the second one will be 0 correct? because it would hit the floor coming to a stop. So once again i have to isolate t, but this time i have the following variables:
Yi = 150 meters
Yf= 0
a=-9.8 m/s^2
Vi= Vf of part 1
In a constant acceleration problem, there are typically five variables: initial speed, final speed, acceleration, distance and time. Given any three you can calculate the remaining two. Accordingly, there are five SUVAT equations, each involving four variables.
For the 150m stage, which three variables are known? Pick the equation that involves those three and time.
But you will anyway be needing to calculate the speed at that altitude too.
 
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  • #5
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Okay, so i used the second equation to to solve for t, using the quadratic formula. so i got that t=1.90793s. I then used the fourth equation to solve for Vf which came out to be Vf=107.238 m/s. So now I am left only with 2 variables, Vf which now becomes Vi since i am now at the second phase where the engine stops. The second one i have is -g which is -9.8 m/s^2. I know I am overlooking something but I just cant figure it out. Unless Δy is the other variable, but I only have it as initial at 150 for phase 2, and I am not told it started from the ground, infact i am kind of told its not, because it is already going at a velocity of 50m/s. So i am not sure what to do
 
  • #6
haruspex
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Okay, so i used the second equation to to solve for t, using the quadratic formula. so i got that t=1.90793s. I then used the fourth equation to solve for Vf which came out to be Vf=107.238 m/s. So now I am left only with 2 variables, Vf which now becomes Vi since i am now at the second phase where the engine stops. The second one i have is -g which is -9.8 m/s^2. I know I am overlooking something but I just cant figure it out. Unless Δy is the other variable, but I only have it as initial at 150 for phase 2, and I am not told it started from the ground, infact i am kind of told its not, because it is already going at a velocity of 50m/s. So i am not sure what to do
It's now at 150m. What will y be when it lands?
 
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  • #7
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Well it would be 0, because it lands right? So -150 would be Δy? this wouldnt make sense though because it would continue to go up for a little bit before it would start to go back down, right?
 
  • #8
haruspex
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Well it would be 0, because it lands right? So -150 would be Δy? this wouldnt make sense though because it would continue to go up for a little bit before it would start to go back down, right?
That doesn't matter, the equations will take care of it. The acceleration is constant from the time the engines stop until it hits the ground. Note that you will get a quadratic equation for that time, but only one solution will make sense.
 
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  • #9
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Oh, I forgot about this, I was able to get it, thanks guys
 
  • #10
Civil_Lady
I know it seems to be cleared up, but a classmate I was working it wasn't able to follow the thread that well, so I'll shorthand the steps below in case anyone else is having the same issues with this.

1) Engines push rocket up from the ground to an altitude of 150 m. You will need to find the final Velocity at that height because that will be the initial velocity used when calculating the rocket at freefall in the 2nd part of this problem.
2) Know that time is the only variable you don't have here - so use the one equation where t isn't included!
3) Remember initial velocity and initial distance is nothing pre-launch.
4) Now, calculate max height of rocket in freefall. Note that the initial Y will be where the rocket engines cut off, and the initial velocity will also be where the rocket cuts off. Also note that you are looking for max height in this problem (aka: where the rocket stops going upwards and starts to fall down). So the velocity during that change is zero (no longer moving up and not yet moving down). Again, plug into the one equation where t is not a variable.


Hope that helps!
 
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