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.

A) 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

2.4

Step-by-step explanation:

To find the time above and below ymax/2 we use the following kinematic formula

`v=v_(0)-gt`

Where

`v` = Final Velocity

`v_(0)` = Initial Velocity

`g` = Gravity

`t` = Time

Notice that we need to first find the values of velocity to solve this equation. To do that we use the following kinematic formula

`v_(1)^(2)  = v_(0)^(2)  - 2gh`

Where

`v_(1)` = Final Velocity

`v_(0)` = Initial Velocity

`g` = Gravity

`h` = Height

Taking h = ymax/2 we get the following formula

`v_(1)^(2)  = v_(0)^(2)  - 2g((y_(max) )/(2))`

To find the value of value of ymax in terms of v0 we use the law of conversation of energy

`KE = PE\\ (1)/(2)mv_(0) ^(2) = mgy_(max) \\ y_(max) = (v_(0)^2 )/(2g)`

Now we can substitute the value of ymax

`v_(1)^(2)  = v_(0)^(2)  - 2g((v_(0)^2 )/(4g))\\ v_(1)^(2)  = (v_(0)^2 )/(2)\\ v_(1) = (v_(0) )/(√(2) )`

Now using the original equation to find time, we input the value of V1 and find time in terms of V0

Assuming final velocity v is 0 (since at the top velocity is zero)

`v=v_(0)-gt\\ t = (v_(0))/(g)`

This gives us total time from bottom to top

To find time from ymax/2 to top we substitute the value of V1

`t_(1)  = (v_(1))/(g)\\ t_(1) = (v_(0) )/(g√(2) )`

The time he is above ymax/2 is nothing more than the difference between t and t1

`t - t_(1)  = (v_(0) )/(g) (1 - (1)/(√(2)))`

To find the required ratio we divide t1 by (t - t1)

`(t_(1))/(t - t_(1))  = 2.4`