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Homework Help: Rocket travelling up and down - kinematics

  1. Jan 16, 2005 #1
    A model rocket has a constant upward acceleration of 40.0m/(s^2) while its engine is running. The rocket is fired vertically, and the engine runs for 2.59s before the fuel is used up. After the engine stops, the rocket is in free fall. The motion of the rocket is purely up and down.

    What is the maximum height that the rocket reaches?

    What will b e the speed of the rocket just before it heats the ground?

    Yes I have looked through the forums for similar posts similar to my question and I have found one https://www.physicsforums.com/showthread.php?t=44459&highlight=rocket
    However that one does include a given initial velocity and my problem does not. I do not know where to start.

    Thank you very much for your help.
  2. jcsd
  3. Jan 16, 2005 #2


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    Homework Helper

    You can use the same equations mentioned in that thread, except you don't need an initial velocity term.

    The base kinematics equation is [tex]s = ut + \frac{1}{2}at^2[/tex]. In your case, you can use [tex]h = \frac{1}{2}(a-g)t^2[/tex]

    where h is the height at time t, (a - g) is the net upward acceleration (taking the opposing effect of gravity into account) and t is the time.

    For the second part, just use [tex]v^2 = 2gh[/tex]. That can be gotten either from conservation of energy or the base kinematics equation [tex]v^2 = u^2 + 2as[/tex]
  4. Jan 16, 2005 #3


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    Science Advisor
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    The problem is really simple.Compute the acceleration upwards.Then use the fact that gravtity does not vary significantly and conclude that the acceleration u computed is constant.
    Then find the height and the final velocity in free fall.

  5. Jan 16, 2005 #4
    Thank you very much for you help everyone

    I have worked it out and got the final answers

    Greatest height = 94.375 meters
    Velocity at ground = -43.0083 m/s

    Can someone please verify?

    [] Greatest Height

    a = 40m/s^2 - 9.8m/s^2
    t = 2.5 seconds
    Velocity Final and Initial = 0
    Xo = 0
    X =?

    Using the position function

    x=1/2(a)(t^2)+Vo(t) +Xo

    = (1/2)(40 - 9.8)(2.5^2)

    [] Velocity at Ground

    a = - 9.8m/s^2
    t = ?
    Velocity Final = ?
    Velocity Initial = 0
    Xo = 94.375
    X = 0

    Using the position function

    0=(1/2)(-9.8)(t) + 94.375

    t=4.3886 seconds

    Plug time into Velocity function (deriv of postion)

    Velocity = -9.8(4.3886)

    = -43.0083 meters/second
    Last edited: Jan 16, 2005
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