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How high does the basketball player leap?

  1. Nov 8, 2013 #1
    1. The problem statement, all variables and given/known data

    From a stationary standing position, a basketball player leaps straight up, reaching a height og 1.0 m above the ground. How long is the player in the air?

    2. Relevant equations

    I think I could use y=y0t+v0t+1/2at^2
    3. The attempt at a solution

    I've found the formula that looks best to use for this problem. I know a=9.8 m/s^2. I know, letting up be positive, that y=1.0 m and yinital=0 m. The equation I want to use wants me to know the inital velocity. How do I find it? I don't think it is zero, because he's leaping up. I could use v^2=v0^2=2aΔx, but I don't know the final velocity, either.

    Thanks in advance for clarification!
     
  2. jcsd
  3. Nov 8, 2013 #2

    CWatters

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  4. Nov 8, 2013 #3
    Thank you!!

    Okay, I get that the acceleration is 9.8 m/s^2, and the displacement is 1.0 m. Is the final velocity 0 since it he stops at his peak of the jump before falling?
     
  5. Nov 10, 2013 #4

    CWatters

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    Sorry for delay in replying. Yes. Final velocity at the top is zero (briefly!)
     
  6. Nov 11, 2013 #5
    Thank you! The delay is quite alright.So I can use 0 as the final velocity for my entire equation?
     
  7. Nov 11, 2013 #6

    CWatters

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    I'm more familiar with it written..

    s = vt - 0.5at2

    where
    s = displacement (height) = 1m
    v = final velocity = 0
    a = acceleration = -g
    t = time
     
  8. Nov 11, 2013 #7
    Okay, thanks! I think there's a compromise here. I only know it well the other way, so I'll work it out, and you can tell me if that step was right. Thanks again!
     
  9. Nov 11, 2013 #8
    If it doesn't add complications, I think I'll let up be positive.
    y=y0t+v0t+1/2at^2
    y=1/2 at^2
    1=1/2 (9.8*t)

    1=4.9*t

    t=0.204 s.

    Is that right? Would I need to double that answer?
     
  10. Nov 11, 2013 #9

    haruspex

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    v0 usually denotes initial velocity, and indeed that is the form of equation you've written. But you don't know v0. However, you can reverse the problem and ask how long it takes to fall from that height, starting at rest. This is effectively what you have done.
    Even so, there is an error. What happened to the power of 2 on t?
     
  11. Nov 11, 2013 #10
    Oh, I forgot to square it. Thank you! I thought the inital velocity was 0, since before he jumped, he wasn't moving. Is it the rate at which he jumped, instead?
    y=y0t+v0t+1/2at^2
    1=v0t+1/2(9.8)^2

    Now do I need a different equation to find the inital velocity?
     
  12. Nov 11, 2013 #11

    haruspex

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    Yes, it's the launch speed. If you set v0=0 and apply the formula the distance you get will be how far you'd fall in that time. As I wrote, that will answer the question, simply by thinking about the process in reverse. But if you want to do it by standard methods then you need an equation, or equations, involving s, t, a and vf. CWatters quoted such. If you want to get there using the equations you know, you will need to use two of them and eliminate v0 between them.
     
  13. Nov 11, 2013 #12
    Okay, thank you! So you're basically saying I need another equation to find the inital velocity, then I can use my equation to find the time, correct?
     
  14. Nov 11, 2013 #13

    haruspex

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    That'll work.
     
  15. Nov 11, 2013 #14
    Okay, thanks! Would v^2=v0^2+2aΔx would work? I think I know all the needed information to solve, but since v^2 equals 0, the equation doesn't seem to be in the right format.
     
  16. Nov 12, 2013 #15

    haruspex

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    Why is that a problem?
     
  17. Nov 12, 2013 #16
    Oh, I just now realized it is a quadriatic equation. Correct?
     
  18. Nov 12, 2013 #17

    haruspex

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    Yes.
     
  19. Nov 12, 2013 #18
    Thanks!

    0=v0^2+2(9.8)(1)

    0=v0^2+19.6

    Now that I simplify, it looks like I only have three terms instead of four. Am I suppossed to forget about simplifying until the end?
     
  20. Nov 12, 2013 #19

    haruspex

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    Watch the signs!
    As I wrote in post #11, you want two equations involving v0 and eliminate v0 between them. It's always best to leave plugging in numbers to the final step. It makes it easier to spot mistakes, easier for others to follow, and minimises accumulation of rounding errors.
     
  21. Nov 12, 2013 #20
    Okay, so this is a quadriatic equation, but I *don't* want to use the quadriatic formula to solve, correct? Thanks again!
     
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