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.

the athlete spends 2.4 times more time at the upper part of his way than in the lower one.

Explanation:

Let's find the velocity V1 of an athlete to reach half of the maximum height equation

V1 = v20 - 2gh = v20 - 2g (ymax) / 2

Here, Vo is the initial velocity of athlete, v1 is the velocity of athlete at half the maximum height, g is the acceleration due to gravity, h=ymax / 2 is half of the maximum height.

We can fund the maximum height that athlete can reach from the law of conservation of energy

KE = PE

1/2M v20 = mg ymax

ymax = v20 / 2g

Then, substituting ymax into the first equation we get

V21 = v20 - v20/2 = v20/2

V1 = V0/∫2, we can find the time that the athlete needs to reach the maximum height (ymax) from the kinematic equation

V = V0 - gt

Here, V is the final velocity of an athlete at the maximum height; V0 is the initial velocity of an athlete

Since, V=0ms-1, we get t=V0/g

Similarly, we can find the time t1 that an athlete needs to reach maximum height from the Ymax/2:

T1 = V1/g = V0/g∫2

So, it is obvious that the time to reach Ymax from Ymax/2 is nothing more than the difference between t and t1:

t-t1 = V0/g (1-1/∫2)

finally, we can calculate the ratio of the time he is above Ymax/2 to the time it takes him to go from floor to that height.

T1/t-t1 = V0/g∫2V0 * g∫2/∫2-1 = 2.4

Answer; the athlete spends 2.4 times more time at the upper part of his way than in the lower one.