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Kicking a Ball Upwards with force x and mass y

  1. Nov 24, 2009 #1

    Gui

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    1. The problem statement, all variables and given/known data
    Problem (this is an extra credit challenge kinda thing): You kick a 2kg ball with 50N of force, directly upwards. The ball is not moving when you kick it. How far does it go and how long is it in the air (take gravity into consideration)?

    2. Relevant equations
    I don't know, but I've been discussing this on #physics at freenode and 2 equation have been proposed although we haven't managed to figure it out yet: a = a0 - g*t AND v = v0 + at
    I'm not sure whether each of them is right or not, but it's been tossed around and discussed. I've also been told I don't have enough information, but my teacher told me I did.

    3. The attempt at a solution
    First, I tried this:

    force = 50N

    mass = 2kg

    acceleration = ? = 25 m/s/s

    gravity = 9.8 m/s/s

    a = a0 - g*t (the second derivative of position equation x = x0 + vt+at^2/2)

    a = 0m/s/s (the acceleration at the top when it stops)

    a0 = 25m/s/s (the initial acceleration i got from f=m*a)

    g = 9.8m/s/s

    t = ?

    0m/s/s = 25m/s/s - 9.8m/s/s*t

    t = 2.551

    Then I was told: "Wait a sec... what are you dealing with change in acceleration for" ... " your equation is incorrect ... your equation has acceleration confused with velocity"

    So I tried this:

    force = 50N

    mass = 2kg

    acceleration = ? = 25 m/s/s

    gravity = 9.8 m/s/s

    v = v0 + at (derivative of vf - vi = at)

    v = 0 (final velocity, at the top when it stops)

    v0 =

    I'm not sure what to do from now on though. Any help is appreciated - thanks! :D
     
  2. jcsd
  3. Nov 24, 2009 #2
    I think you don't have enough information.
    The crucial piece of the puzzle that's missing is how long the 50N force was applied? 0.01 seconds? 0.1 seconds? It makes a difference.
    Without that, you can't determine the change in momentum, from which you may deduce the starting velocity v0. Determining the change in momentum:
    F*t = m2*v2 - m1*v1
    t being the time of contact and in this case m1=m2, v1=0.

    From there on there are several methods to solve this, a simple one is using energies.
    1/2 m v^2 = mgh
    starting position's kinetic E = potential E of pause at the top

    once you got the hight, you can calculate time from 1/2 a t^2.


    Best wishes
     
    Last edited: Nov 24, 2009
  4. Nov 24, 2009 #3

    Gui

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    Yeah, I got told that too... so if I just assume the amoutn of time the foot is in contact with the ball to be 0.1s, then I can do it. Thanks for your input, I really appreciate it :)

    I'll try and solve it now and I'll post back results
     
  5. Nov 24, 2009 #4

    Gui

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    I got 2.5m/s.

    I don't quite understand this part though, can you elaborate a little? What do the variables stand for in that equation? Is the equation just a derivative of ke = 1/2mv^2? Also, what do you mean with last line?
    Thanks.
     
  6. Nov 24, 2009 #5
    If you don't know what potential energy or kinetic energy is, just solve it with your original idea. Also read http://en.wikipedia.org/wiki/Kinetic_energy and http://en.wikipedia.org/wiki/Potential_energy :smile:

    Otherwise...
    kinetic energy + potential energy = constant
    because energy is reserved within a closed system. (Sorry if that's not the proper expression, my English isn't the best.)

    At the starting position (after the kick) for a moment at the very beginning the ball only has kinetic energy.
    At the top the ball stops to move for a moment, thus it has 0 kinetic energy. But energy can't just be lost (assuming no air resistance, etc), so what happened to it? It changed into another form of mechanical energy, potential energy: m*g*h m=mass g=gravity h=height.

    The two positions:
    bottom: 0 + kinetic
    top: potential + 0

    state 1) = state 2) because energy is reserved within the system.

    Once you have the height, you may calculate the amount of time the ball is in the air. Think of how long it would take to drop the item from that height (h). And don't forget to multiply that by two (going up takes the same time as going down).
     
    Last edited by a moderator: Apr 24, 2017
  7. Nov 24, 2009 #6

    Gui

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    I understand it, thanks mate! :D
     
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