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Hang Time for Vertical Leap

  1. Oct 25, 2014 #1
    The maximum "hang time" for a human who jumps in the air under his own power is said to be less than 1 second. This includes jumping on the spot, running jumps, hops, leaps, dives, and bounds. Javier Sotomayor (Cuba) is the current men's record holder with a jump of 2.45 m (8 ft 1⁄4 in) set in 1993. Using a stop watch, I timed it (see below) at less than 1 second. So it would appear that the 1 second rule holds in this case, and I'd wager even Michael Jordan couldn't stay in the air any longer.

    If H = 2.45 m, the total vertical distance would be 2H = 4.50 m. H = 1/2 gt^2, where g = 9.81 m/sec^2. So t = sqrt(2H/g) = sqrt(4.9/9.81) = 0.71 seconds. Still less than a second in the air!

    I did a bit of calculating and found that a person would have to execute a vertical leap of about 11 feet 4 inches (~16/sqrt2 feet) or 3.47 m (~4.9/sqrt2 meters) in order to stay off the ground for one second. This is the time elapsed from when the last part of your body leaves the ground to when the first part of your body touches down.



    Does anyone else get the same results?
     
  2. jcsd
  3. Oct 25, 2014 #2
  4. Oct 25, 2014 #3
    So even Michael Jordan has a maximum hang time of 0.92 seconds, and that's with some horizontal momentum converted to vertical momentum. The <1 second rule still rules.
     
  5. Oct 27, 2014 #4
    I have to correct myself on this. If H = 2.45 m, it would be time going up = time going down. So this would be the same as twice the time it takes an object dropped from a height of 2.45 m to reach the ground. t = 2*(2*2.45 m / 9.81 m/sec^2)^0.5 =
    1.413 seconds. So greater than 1 second. But this is not a vertical leap from a stationary position. Some of the horizontal momentum is transferred to vertical momentum.

    For a "hang time" of 1 second from a stationary position, d = (9.81*0.5^2)/2 = 1.23 meters = 4.02 feet (48.27 inches).

    Glad we got that cleared up!
     
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