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Rocket's maximum height

  1. Jun 16, 2003 #1
    I have to take a final soon and I have practise questions The only one here I am having trouble with is thislast one. This question bugs me and the answer key doesn't show how to do it

    After a model rocket reached its maximum height, it then too 5.0 seconds to return to the launch site. What is the approximate maximum height reachd by the rocket? {Neglect air resistance.}

    the answer is 120 m

    the only info u have is time and acceleration

    How do I go about doing it?
  2. jcsd
  3. Jun 16, 2003 #2
    the rocket going up is confusing, but i think this is what you're looking for:
    simply, D=RT (distance=rate x time, if you're not familiar with it)

    my guess is that this is not correct becuase you have to take into account the return of the rocket because of gravity. so something you muliply by either 2 or 1/2. maybe...
    Last edited: Jun 16, 2003
  4. Jun 16, 2003 #3


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    The question asks: how far does something fall in 5 seconds, when its acceleration (due to gravity) is a = 9.8 m/s^2?

    s(t) = 1/2 a t2, where t = 5, yields 122.5 m.

    - Warren
  5. Jun 18, 2003 #4


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    Maximus, "D= RT" only applies when R is a constant.

    If R is changing at a constant rate (that is the rate of change of R is a constant) then you can use an "averaging" method. When the rocket was at it's peak, it's speed was 0 (that's why it stopped going up!). After 5 seconds at a constant acceleration of -9.8 m/s^2, it's speed is -9.8*5= -49 m/s (Since the rate of change of speed, acceleration, is constant, you CAN use "RT"). The average of the two values is (0+(-49)/2= -24.5. Using that average value, in 5 seconds, the rocket will fall -24.5*5= -122.5 m. The rocket must have fallen from a height of 122.5 m, value chroot (Warren) gave.

    Using "g" instead of -9.8 m/s^2 and t instead of 5 seconds, the two speeds are 0 and -gt so the "average" speed is (0-gt)/2= -gt/2.
    Multiplying that by t to get distance gives (-g/2)t^2, the formula chroot used.

    Again, this only works when the acceleration is constant (a very important special case!). If the acceleration (in general "rate of change of the rate of change") is not constant, then you will have to learn calculus!
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