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Homework Help: Free fall of a rocket

  1. Feb 1, 2008 #1
    This free fall problem seems simple, but I can't get it.

    you have a rocket fired from rest with an acceleration of 40 m/s^2. It runs for 2.5 and then falls.

    I'm wanting to find the maximum height, which I think would require me getting the initial velocity first.

    v = v0 + at
    0 = v0 + 40(2.5)
    v0 = -100 (wrong, it's going up?)

    then I thought just plugging into the equation x = x0 + v0t + .5at^2 would work, but the v0 is obviously wrong. I don't see what I'm doing wrong, though. The answer should be 635m.

    Can anyone point out where I'm going wrong?
  2. jcsd
  3. Feb 1, 2008 #2
    [Post Deleted]
  4. Feb 1, 2008 #3
    Vo of the rocket is zero at time 0 and at all times for the equations used while it is accelerating to it's maximum velocity. It will reach it's maximum velocity in 2.5 seconds. Knowing this you should solvee for the velocity at time = 2.5 second(remember the total acceleration on the rocket is not just the 40 m/s*s but also -9.8 m/s*s from gravity). You should also solve for the distance traveled in 2.5 seconds (about 94.4 meters). After the 2.5 seconds the rocket is in free fall with an acceleration of -9.8 due to gravity but with a positive velocity of about 75.5 m/s. Question to answer now is how long will it take before the velocity = zero? Once you have that time solve for the additional height the rocket will travel. The total height of the rocket will be in the area of 385 meters. Hope this helps.
  5. Feb 1, 2008 #4
    since it has 0 initial velocity and assuming it starts at the origin, its height once it runs out of gas is @ 250m , by using y = 0 + 0 + 40(2.5^2)

    THIS SOLUTION is assuming that the rocket went up with exactly an accell of 40m/s^2,, not 40m/s^2 -9.8.
    the question isnt clear to me so thats that.
    the reason i assumed this was because it didnt say that it launches with an initial velocity of 40m/s and then gravity acted upon onit, it said that it launched with an acceleration of 40m/s, am i wrong? acceleration caused by gravity is constant and is always acting,,,therefore how can you launch at 40m/s^2 and THEN factor in gravities 9.8.
    so constant acceleration till t = 2.5 , of 40.

    its velocity once it runs out of gas must be 100m/s, i think...just 40 * 2.5

    then use
    v^2 = v0^2 + 2a(y-y0)
    velocity at max height is 0
    0 = 100^2 -19.6(y-250) ; solve for y
    y = 760.2

    thats what i got, dont trust me
    im no expert =)
    Last edited: Feb 1, 2008
  6. Feb 1, 2008 #5


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    You should assume that the acceleration going up is 40m/sec^2. Don't subtract 9.8m/sec^2. They GAVE you the acceleration. After it stops accelerating what is it's velocity and height? Now treat these as initial conditions for the next phase of the flight while it climbs to maximum height. During this phase the acceleration is -9.8m/sec^2.
  7. Feb 1, 2008 #6
    From you statement of the problem, I assumed (sorry) that the acceleration of 40 meters per second per second was the acceleration supplied by the rocket motor. Assumming up is positive then the rocket motor would apply 40 m/s^2 up while gravity would supply -9.8 m/S^2 due to gravity. From this I said that the total acceleration on the rocket then is the sum of both accelerations, ie, 40-9.8 = 30.2 m/s^2. If the 40 m/s^2 acceleration is the total acceleration on the rocket including the affect of gravity, then the problem is easier to solve. The equation is V(max after 2.5 seconds) = 0 + 40*2.5 = 100 m/s.

    Now what is the total distance traveled in 2.5 seconds when the initial velocity is zero, the acceleration is 40m/s^2. s= 0.5*40m/s^2*2.5s*2.5s = 125 m. At this point the rocket no longer has a positive acceleration of 40 m/s^2 but a negative -9.8 meters/s^2. It also has a positive velocity now of 100m/s and will continue to climb until it has lost this velocity or velocity = 0. At this point the rocket will be as high as it can go before falling back to earth. USing the velocity equations then to get the time it will take to reach zero velocity. 0 = 100 m/s - (9.8m/s^2)*t. Then solve for the additional climb distance with s=0.5(a*t*t)+(100m/s)*t. Add this to you height at 2.5 seconds to get a total height.
  8. Feb 2, 2008 #7
    Thanks to everyone who posted. 40 m/s was the total acceleration and after reading the posts my biggest problem was assuming that 2.5 s was the time it took to reach the top.
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