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Model rocket

  1. Feb 4, 2005 #1
    A model rocket is launched straight upward with an initial speed of 45.6 m/s. It accelerates with a constant upward acceleration of 1.5 m/s^2 until its engines stop at an altitude of 180.5 m. What is the maximum height reached by the rocket? Answer in units of m.

    How long after lift-off does the rocket reach its maximum height? Answer in units of s.

    How long is the rocket in the air? Answer in units of s.
     
  2. jcsd
  3. Feb 4, 2005 #2
    Ok, do you have a kinematics formula sheet. I can't just keep telling you what formulae to use... :surprised
     
  4. Feb 4, 2005 #3
    lol i have a physics book but i dont know what section it is or what to look up
     
  5. Feb 4, 2005 #4
    Here are the basic formulae:

    [tex]V_f^2 = V_i^2 + 2ad[/tex]

    [tex]d = V_it + \frac{1}{2}at^2[/tex]

    [tex]d=vt[/tex]

    [tex]V_f = V_i + at[/tex]
     
  6. Feb 5, 2005 #5
    This problem is based on fundamental equations of motion as mentioned by christinono.
    In this case Vi=0,a=-9.81m/s^2 and other quantities can be directly substituted to get the desired result.
     
  7. Feb 5, 2005 #6
    I can see how Vi=0 because its a constant acceleration, but how do you calculate the difference when the engines shut off? and how much farther it goes. i got 51.179 as my Vf, then do i subtract -9.8 from that and get the time after it that it takes then have that be my total time? and distance?
     
  8. Feb 5, 2005 #7
    Where did you get Vi = 0 /ms? Vi is 45.6 m/s, as the question says. Rather, the final velocity is zero, since it's asking for the MAX height. At that point, it will stop for a split second and come back down. All you have to do is look at all the formulae you have and decide which one will allow you to find the unknown with the info you are given. In this case, you will have to split it up into 2 steps.
    1) Find the final speed of the rocket (by final, I mean when the engine stops). You know:
    Vi = 45.6 m/s
    [tex]a = 1.5 \frac{m}{s^2}[/tex]
    V2 = ?
    d = 180.5 m

    2) The value you found for Vf in 1) will be the initial velocity for the second part. Now, you wanna find the distance travelled by the rocket once the engine stops. Here's what you know:

    Vi = whatever you found in 1)
    [tex]a = -9.81 \frac{m}{s^2}[/tex]
    V2 = 0 m/s
    d = ?

    Get it?
     
  9. Feb 5, 2005 #8
    yea i got all that right, just the problem im having now is the total time. i thought it would be 3.729557083(engines)+5.2239111790(engines off)+ (314.217691/9.8)(free fall)=41.01649796 total time?
     
  10. Feb 5, 2005 #9
    Is that not the answer in the book. If not, show me your work, and I'll tell you what you did wrong.
     
  11. Feb 5, 2005 #10
    in question 26 it asks how long does it take to reach max h, i put in 8.95347 s and that was correct then for 25 the max height in m, i put in 314.217 and that was correct, then 27 is total time in air so i did 314.217/9.8 (no other indication of gravity) and got 32.06265305 and then added the answer from 26 to it the 8.95347 and got 41.0162406 s but when i typed that it in said incorrect
     
  12. Feb 5, 2005 #11
    Let me make sure I understand:
    You didn't get the right answer for the total time the rocket was in the air and you found that the max height was 314.217m (which was right)?

    If that's the case, you need to use a kinematics equation to find the time for the rocket to come down once it has reached its max height. You know the following:
    Vi = 0m/s
    [tex]a = -9.81 \frac{m}{s^2}[/tex]
    d = 314.217m (if that's right)
    t = ?

    Can you find a formula that fits?
     
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