Posted by bob on .
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,
You have it correct, I can't help you with your computer acceptance format. Maybe they want 1/sqrt2 I have no idea.

physics 
bob,
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,
How did you figure this problem out? How did you work it out?

physics 
Luis,
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 
physics 
James Duk,
jesus

physics 
Alex,
The answer was 4.82. I tried solving it, but I ran out of attempts inputting .707,etc. I still have no idea how