Question

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.

Answer 1

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

Step-by-step 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.