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Kinematics free fall

  1. Oct 31, 2010 #1
    1. The problem statement, all variables and given/known data
    . A ball is thrown straight up at t=0. It is found to be falling with a speed of 8 m/s at t= 6 sec. Find the displacement of the ball.


    2. Relevant equations



    3. The attempt at a solution
    v=v0-gt
    8=v0-60
    v0=68
    deltax =v0t - .5gt2
    delta x = (68*6)- 5*36=228
    but it says the answer is -132 am i wrong or is it wrong?

    Edit:
    nvm I got it it was supposed to be -8
    -8=v0-60
    52=v0
    deltax=52*6 - .5*10*36 = 132

    uhm it seems I messed up again this time I got a plus why isn't it a minus?
     
    Last edited: Oct 31, 2010
  2. jcsd
  3. Oct 31, 2010 #2
    Also I have another question is delta y = v(t)t+.5 gt^2 considered one of the kinematics equations I mean can I use it without having to derive it every time?
     
  4. Oct 31, 2010 #3
    "Find the displacement of the ball."

    Since you throw the ball straight up and it comes back down to where you initially threw it from, then I believe the displacement should be zero. It would be like running a lap around the track. Sure you have ran a 1/4 of a mile but your total displacement would be zero because you end up back to where you originally started.

    "Also I have another question is delta y = v(t)t+.5 gt^2 considered one of the kinematics equations I mean can I use it without having to derive it every time?"

    Yes that is one of the kinematic equations that you can memorize. It only applies, however, to constant accelerations.
     
  5. Oct 31, 2010 #4
    i am pretty sure it meant the displacement at t=6 ...
     
  6. Oct 31, 2010 #5
    Did you use a positive acceleration or a negative acceleration? Always remember which way you set up your coordinate system (i.e. down is negative and up is positive.) If you used (-8)m/s for the velocity than we assume down is negative.
     
  7. Oct 31, 2010 #6
    I used negative acceleration. - 10
     
  8. Oct 31, 2010 #7
    displacement = x(final)-x(initial)
    s = 0-132
    When you used 52 m/s (which was your initial velocity) for your motion equation, you get an initial position.
     
  9. Oct 31, 2010 #8
    x final is 132 and x initial is 0 why did you reverse it ?
     
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