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Homework Help: Free Fall Question

  1. Jan 28, 2008 #1
    A particle is launched vertically at an upward velocity of 39.24m/s. Neglecting air resistance
    what is the paricle's velocity on impact
    average acceleration
    average velocity
    and displacement after impact
  2. jcsd
  3. Jan 28, 2008 #2
    What equation do you think you will need for this question?

    Also, you need to show some work.
  4. Jan 28, 2008 #3
    0=39.24^2 +2(-9.8)s

    vf^2= 1539.78+19.6(157.12)
    vf= 67.96m/s
    displacement = zero
    average acceleration = (28.72)/2.9 = 9.8m/s ------right??
    If the displacement is zero then how do you find the average velocity??
    Last edited: Jan 28, 2008
  5. Jan 28, 2008 #4
    Since it is a straight free fall v0=39.24 vf=-39.24
    t=8.00 sec

    avg Acceleration= -78.48/8=9.81
    avg velocity = 0-0/8 = 0???????
    displacement = 0

    Any help???
  6. Jan 28, 2008 #5
    Do you use vf=-v0
    or do you use vf^2= 1539.78+19.6(157.12)
    where 157.12 is the greatest height times 2

    Which do we use?
  7. Jan 28, 2008 #6
    0=39.24^2 +2(-9.8)s

    Why did you assume that the final velocity would be zero? Especially when the problem ask you what the final velocity would be on impact.
  8. Jan 28, 2008 #7
    was the maximum height.

    Do you use vf=-v0
    or do you use vf^2= 1539.78+19.6(157.12)
    where 157.12 is the greatest height times 2

    I think it is vf=-v0
  9. Jan 28, 2008 #8
    If you do use vf=-v0 then vf=-39.24m/s
    then avg acceleration= -39.24-39.24/8=9.8m/s

    displacement is 0 because the particle went up then down.
    How do you find the avg velocity = delta s/delta t
    when delta s = 0?????????
  10. Jan 28, 2008 #9
    to find avg velocity do I use the distance which is the max height times 2 or do I use the displacement of 0??????
  11. Jan 28, 2008 #10
    I have to turn this in tomarrow. So any help on whether to use the displacement of 0 or the distance and the time to find the avg velocity would really be appreciated.
  12. Jan 29, 2008 #11
    for upward motion:
    u= 39.24 m/s
    >s= 78.56 m

    now for downward motion:



    this is the velocity after impact.

    avg accln=9.8 m/s^2

    avg velocity=0 [ since displacement=0]
  13. Jan 29, 2008 #12
    Is that velocity on impact because that is what I need to find
  14. Jan 30, 2008 #13
  15. Feb 1, 2008 #14

    Shooting Star

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    Homework Helper

    One small tip for future use. The rise and fall of a particle from and to the same point is symmetrical. That is to say, the time of rise is equal to the time of fall. It has the same speed going upward at a certain point as when crossing the same point downward again. No wonder the velo of impact equals the velo of throw. The time taken to rise from h1 to h2 in between is the same as the time to fall from h2 to h1 when coming down.

    Of course, you must deduce all these at least once in your life, to make life easier thence.
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