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Model rocket physics problem

  1. Sep 5, 2004 #1
    A model rocket has a constant upward acceleration of 40.0 m/s^2 while its engine is running. The rocket is fired vertically, and the engine runs for 2.50 s before it uses up the fuel. After the engine stops, the rocket is in free fall. The motion of the rocket is purely up and down. i have two questions.

    1.) What is the maximum height that the rocket reaches?

    ok first i tried to find the average velocity for the up trip. 40/1.25 = 32
    and after i found 32, i did 32*2.50s = 80 which is the wrong answer.

    2.) What will be the speed of the rocket just before it hits the ground?

    and for #2, i dont know how to start it.
  2. jcsd
  3. Sep 5, 2004 #2


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    The average speed on the way up is [itex]40\times2.5/2[/itex].
  4. Sep 5, 2004 #3
    For no 2, you use the maximum height that you've calculated in the equation v^2 = u^2 + 2as, using a = g. The initial velocity, u = 0 (obviously, as otherwise it wouldn't have reached the maximum height).
  5. Sep 5, 2004 #4


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    "ok first i tried to find the average velocity for the up trip. 40/1.25 = 32
    and after i found 32, i did 32*2.50s = 80 which is the wrong answer."
    Sinjce there is a constant accleration, the velocity at the end of of the 2.5 seconds is 40*2.5= 100 m/s so the average velocity is 50 m/s (you multiplied by 2/t rather than t/2!). After 2.5 seconds, the rocket has a height of 2.5*50= 125 meters.

    However, that is just the height when the engine shuts down. It has upward momentum so it continues upward.

    After the engine shuts off, the only acceleration is that due to gravity, -9.8 m/s2. After t seconds, the speed will be 100- 9.8t (remember that the rocket's speed at engine shut off was 100 m/s). As long as that is positive, the rocket will continue going higher. When it is negative, the rocket will be coming down. The highest point is when the speed 100- 9.8t= 0. You can solve that for t to find how long the rocket continues to "coast" upward. Also, since the rocket's initial speed at engine shutdown was 100 m/s and at the end is 0, you know that the rocket's average speed during coasting was 50 m/s (again, constant acceleration). Multiply that by the time the rocket coasted upward to get the height gained during that time. Don't forget to add the 125 m to that to find the maximum height.

    It's going to be a heck of a lot more than 80 m!

    There are several ways to answer (2). One way would be to find the potential energy at the highest point (mgh) and set that equal to the kinetic energy of the rocket just as it hits the ground ((1/2)mv2). You don't know m but that's alright, they will cancel.

    Another way is to imagine that rocket gained all of its speed instantaneously (like the "throw a rock upward" problem). If the rocket had speed v0 at the first instant, then its height at time t would be given by -(g/2)t2+ v0t. The time to the highest point would (until its speed is 0) would be given by -gt+ v0= 0 or
    t= v0/g. Putting that into the height equation, the greatest height would be -(g/2)(v02/g2)+ v0(v0/g= v02/2g (looks a lot like the energy equation doesn't it?) Set that equal to the height you got before and solve for v0. That would be the "initial speed" but because the motion is symmetric, will also be the speed with which the rocket hits the earth.
    Last edited: Sep 5, 2004
  6. Sep 5, 2004 #5
    ah perfect, and i totally get what your saying. thanks for the help
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