Seems simple enough but

  • Thread starter JinFTW
  • Start date
In summary: Also, by 10 seconds the velocity is already -438 m/s, which means the rocket has already fallen to an elevation of 438 m (or 1810 m - 438 m = 1372 m above the ground). So when the engines shut off, the rocket is still moving upwards, but it will soon start moving downwards because the acceleration is still negative. Once it starts moving downwards, the velocity will continue to become more and more negative, until it hits the ground again. In summary, the conversation discusses a physics problem involving a stolen rocket with a time-dependent thrust and a mass of 2kg. The main issue is finding the rocket's maximum elevation and time airborne, with the added complication of a non-constant
  • #1
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Hey guys, I just joined the forum and I hate to bother you with this, but after working on my HW questions for awhile I've found one that has completely killed my brain. No matter what I do I cannot get an answer that seems suitable to what the question is asking. Any help on this would be greatly appreciated. So the question is as follows:


The troublesome Mooninites Ignignokt and Err have stolen a rocket from
NASA and escaped to Mars. On Mars, Err opens the classified instruction
documents and learns that this is no ordinary rocket but one with an experimental
engine that provides a time-dependent thrust for exactly 10 seconds after which the
engines shut off completely. The thrust provided by the rocket engines can be
expressed as F(t) = −24t 2 +144t
and is purely in the vertical direction. Up to no good as usual, they decide to fire off
the rocket from the surface of Mars. Neglecting air resistance and assuming the
rocket has a mass of 2kg, how long is the rocket airborne before it crashes back to
the ground and what is the rocket’s maximum elevation?

The Thrust equation (having a negative involved) is messing everything in my equation up. According to my professor this problem should only take me five minutes, but I've spent the last two hours doing velocity and accleration for each and every second as well as for the problem itself, and nothing is getting me anywhere. Any assistance in helping me keep my sanity would help a ton! Thanks for you time guys.
 
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  • #2
Sorry guys just read the rules: Ok so I started this problem by using
F = T (thrust) - mg (weight) = ma

Since m = 2 this means that I simply have to plug in t into the Thrust equation to get the force of the thrust.

for example under 1 second it's:
F = 120 - 2(3.8) = 2a

which means a = 56.2 m/s ^2 at 1 second

I continued doing this sort of thing for 1 - 10 seconds. My plan from here was to split the height part of the problem into two parts:
First, to find the height till the engines quit (thrust stops) and then to find the height from there till final velocity = 0.

So I use the formula
Yf = Yi + VoT + 1/2at^2
Since Yi and Vo = 0, this becomes
Yf = 1/2at^2

For time I used 10 seconds
For acceleration I originally used 56.2, but then I tried a from 10 seconds, which came out to -483.8 m/s ^2 (seems really wrong). What upsets me here is that acceleration is not constant on this problem, yet my class has never HAD a problem without constant accel, therefore I'm not entirely sure which accel I'm supposed to use.

I DO know that once I find out Yf, I can do the second part since I know a in that case will be (3.8 m/s^2)

Therefore for the second part of height you use:
Vf^2 - Vi^2 = 2a(Yf - Yi)
 
  • #3
I don't see how you can solve this without calculus. But if you've never had a problem without constant acceleration, then it seems unlikely to me that you know calculus. For what it's worth, in case my assumption is wrong, here's how I'd approach it:

Let's assume some initial conditions

v(0) = 0 (the initial velocity is zero)
y(0) = 0 (the initial position is zero)

The rocket is at rest on the ground when the engines fire.

[tex] a(t) = -12t^2 + 72t - g [/tex]

[tex] v(t) = \int a(t) \, dt = -4t^3 + 36t^2 -gt +C_1 [/tex]

[tex] C_1 = v(0) = 0 [/tex]

[tex] y(t) = \int v(t) \, dt = -t^4 + 12t^3 -\frac{1}{2}gt^2 +C_2 [/tex]

[tex] C_2 = y(0) = 0 [/tex]

y(10) = 1810 m --> this is the height when the engines shut down
v(10) = -438 m/s --> this is the velocity when the engines shut down

Now that the engines are off, the acceleration is just g downward and this has changed into a free fall problem with initial conditions y(10) and v(10). The time airborne is just 10 seconds + the time it takes for an object to free fall 1810 m on Mars. We can calculate the latter using the new equation governing the trajectory:

[tex] y(t) = y(10) +v(10)t - \frac{1}{2}gt^2 [/tex]
 
  • #4
Thank you very much for the help, but I was just curious, I can see how you got the equations to get Vf for the first part of the rocket's trip. What I'm curious about is the last part of your equation. Are you saying that this rocket doesn't keep going after thrust ends? I know I've done a rocket problem before (not as complicated as this one) and in finding the max height I first had to find when the engines quit, and then from where the engines quit till Vf = 0. If Vf for the first part is negative before the engines even quit, how does it keep going up?
 
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  • #5
Not sure if this is the final answer or not, but for time I have (10s + 4.06s) = 14.06s. I also am trying to figure out how Yf = 1810 (if it does), even though I understand how you got your equations, I've never seen a problem similar to this where we don't have to find where Vf = 0 (top of the range) in order to find max height and total time.
 
  • #6
JinFTW said:
If Vf for the first part is negative before the engines even quit, how does it keep going up?

The rocket had already started moving downwards by 10 s. When I plotted the acceleration, velocity and position vs time, I found that the max height was > 1810 m, it occurred at around 9 s, and it was at that point that the velocity was zero. By 10 seconds, the velocity was negative, and the rocket had already moved down to 1810 m. This is because the acceleration is not constant...it is time-varying and it is already negative by the 10 second mark.
 
  • #7
Yeah, it seems strange that the engine thrust goes negative after 6 seconds, but that's the equation you were given.
 

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