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Hang Time

  1. Sep 13, 2006 #1
    Here with a question, and I'm a bit crunched for time. I'm also here to learn, too.
    A basketball player, standing near the basket to grab a rebound, jumps 76.0 cm vertically. How much total time does the player spend (a) in the top 15.0 cm of this jump and (b) in the bottom 15.0 cm? Does this help explain why such players seem to hang in the air at the tops of their jumps?

    Thanks in advanced!
  2. jcsd
  3. Sep 13, 2006 #2
    soooo... do you have nay thoughts on what to do??
  4. Sep 13, 2006 #3
    Not exactly, I guess that's why I'm here. I assume the player will be in his top 15.0 cm for a longer time than the bottom 15.0 cm, since the player has to slow down when he's in the air, reverse his direction, and speed up again (downward).
  5. Sep 13, 2006 #4
    can you find the time he will be spending up there, though

    can you find his velocity while he is going upward as he enters the upper 15cm?? can you find the velocity when he is coming down thru the 15 cm again??
    Hint: what is the acceleration?
  6. Sep 13, 2006 #5
    I'm not sure that I can answers those. The acceleration is not constant though I know, seeing as it changes direction at some point. Other than that, I'm just about clueless.
  7. Sep 13, 2006 #6

    who is causing the acceleration?? and which way does it point?

    make sure you are not confusing 'acceleration' with 'velocity'

    still confused?.. what does acceleration do then??
  8. Sep 13, 2006 #7
    The acceleration would be caused by the initial jump upwards? And then gravity pulling him back down?
    Acceleration is increasing speed, right?
  9. Sep 13, 2006 #8


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    Acctually acceleration is CHANGING velocity.
  10. Sep 13, 2006 #9
    OK then, but I'm still not sure what to do next.
  11. Sep 14, 2006 #10


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    With constant acceleration -9.8 m/s2, initial velocity v0, height at time t seconds is x(t)= v0t- 4.9t2 and velocity at time t seconds it v(t)= v0- 4.9t. He will keep going up as long as v(t) is positive, then come back down. That means that he will get to his highest point when v(t)= 0 v0- 4.9t or when t= v0/4.9. Put that into the equation for his height then:
    x(t)= v0(v0/4.9)- 4.9(v0/4.9)2= 76 cm and you can solve for v0.
    Once you know that, you can find the time it takes him to go from x= 0 to x= 15 (the low 15 cm) and the time it takes him to go from 76-15= 61 cm to 76 cm (the high 15 cm). O course, the time he stays in the "high 15 cm" region will be twice that- up to 76 cm then down to 61 cm again.
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