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Kinematics how high does it go problem. Check

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data
    A toy rocket, launched from the ground, rises vertically with an acceleration of 33 m /s^2 for 13 s until its motor stops
    The acceleration of gravity = 9.8 m/s^2
    Disregarding any air resistance, what maximum height above the ground will the rocket achieve? Answer in units of km.

    G = -9.8 m/s^2
    A = 33 m/s^2 with T = 13 seconds

    Unknown
    x - xi = ?



    2. Relevant equations
    x - xi = vi * t + 1/2 A T^2

    V = A * T

    V = Vi - gt




    3. The attempt at a solution

    Ok, so first I broke this problem up into two pieces, I checked the distance it traveled while accelerating by using the first equation I provided... the Vi * t = 0 since my Vi was 0 and the rest I got --- 1960.4 m

    then I saw what velocity it was going at that time and I got 301.6 m/s

    So then I needed to find the distance it traveled from 301.6 m/s to 0 m/s so I used the third equation and set V = 0 to find T which equals 30.7755102 seconds. Then I reused the 1st equation to find the distance of this part

    By plugging g in for my acceleration component of course, I got 4,640.946939 m + my first part 6,601.346939 m / 1000 = 6.601.34 km


    However this answer came up as wrong, ( I still have 6 tries left!) I can't see where I went wrong? I believe it's in the 2nd part finding how much distance the velocity goes until it reaches 0. Thanks for your help!
     
  2. jcsd
  3. Sep 14, 2008 #2

    LowlyPion

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    How fast was the rocket going when the engines stopped?
     
  4. Sep 14, 2008 #3
    I calculated it to be 301.6 m/s

    This is the only problem I'm currently missing but I have 6 more tries and it's due at 11:55 pm please help soon if you wish :D
     
  5. Sep 14, 2008 #4

    LowlyPion

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    What equation did you use for that. Because you know that it accelerated 33m/s2 for 13 seconds. How is that only 301.6m/s?
     
  6. Sep 14, 2008 #5
    Oh, I took 33 - 9.8 which is gravity to get my net acceleration for the time intervaql 13 seconds.
     
  7. Sep 14, 2008 #6
    PROBLEM SOLVED:

    LOL! Thanks, I solved it without subtracting gravity from my acceleration given, that's a poorly worded question though.

    I got 99.85% on my physics homework :D I'm at UT Austin physics 1 btw.
     
  8. Sep 14, 2008 #7

    LowlyPion

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    When they give acceleration that's all you need. When fuel runs out then gravity kicks in.

    Good luck.
     
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