Posted by **bob** on Thursday, October 21, 2010 at 3:23pm.

In the vertical jump, an Kobe Bryant starts from a crouch and jumps upward to reach as high as possible. Even the best athletes spend little more than 1.00 s in the air (their "hang time"). Treat Kobe as a particle and let ymax be his maximum height above the floor. Note: this isn't the entire story since Kobe can twist and curl up in the air, but then we can no longer treat him as a particle.

To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 moving up to the time it takes him to go from the floor to that height. You may ignore air resistance.

NOTE: I got something like 70.7% of the time, but no matter how i write the answer it tells me that I'm wrong. I tried .700, 70.0, and .707 and none were good. PLEASE HELP =]

- physics -
**bobpursley**, Thursday, October 21, 2010 at 3:26pm
You have it correct, I can't help you with your computer acceptance format. Maybe they want 1/sqrt2 I have no idea.

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**bob**, Thursday, October 21, 2010 at 3:35pm
it was a stupid format and i ran out of tries. Basically it wanted me to divide .707 by .293. The worst part is that it was worth a lot =[ o well

- physics -
**Alex**, Thursday, October 21, 2010 at 10:38pm
How did you figure this problem out? How did you work it out?

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**Luis**, Sunday, February 3, 2013 at 2:03am
The time he takes to go up from y(max) /2 to the y{max } is the same as the time he falls from y [max ] to y [ max] /2.

Let y max = H and then y max = H/2

t1^2 = 2g H/2. ===========1

Simnilarly,

Time it takes him to go from the floor to that height

t2^2 = 2g H======================2

1divided by 2 gives

(t1/t2)^2 = 1/2 = 0.5

t1/t2 = 0.707 neaarly 70 %.

t1 = 70% t2.

Thus, for 70% of the total time of going up, he is in the upper half of the total height.

Hence he seems to hang in the air

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