How Long Can a Human Stay Airborne in a Vertical Jump?

[fizixfan]

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?

 

[zoki85]

 

[fizixfan]

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.

 

[fizixfan]

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?

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!

 

[pmacias]

I can confirm that your calculations are correct. The maximum hang time for a human jumping under their own power is indeed less than 1 second, as seen in the record set by Javier Sotomayor. This is due to the laws of physics, specifically the acceleration due to gravity, which dictates the time it takes for a body to return to the ground after being launched into the air. Your calculations accurately show that in order to stay off the ground for one second, a person would need to execute a vertical leap of about 11 feet 4 inches. This is a remarkable feat and it is unlikely that even athletes like Michael Jordan could achieve it. However, it is important to note that other factors such as air resistance and the strength and technique of the individual can also play a role in the hang time of a vertical leap.