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.

Respuesta :

Answer:

2.4

Explanation:

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

[tex]v=v_{0}-gt[/tex]

Where

[tex]v[/tex] = Final Velocity

[tex]v_{0}[/tex] = Initial Velocity

[tex]g[/tex] = Gravity

[tex]t[/tex] = 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

[tex]v_{1}^{2}  = v_{0}^{2}  - 2gh[/tex]

Where

[tex]v_{1}[/tex] = Final Velocity

[tex]v_{0}[/tex] = Initial Velocity

[tex]g[/tex] = Gravity

[tex]h[/tex] = Height

Taking h = ymax/2 we get the following formula

[tex]v_{1}^{2}  = v_{0}^{2}  - 2g(\frac{y_{max} }{2})[/tex]

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

[tex]KE = PE\\ \frac{1}{2}mv_{0} ^{2} = mgy_{max} \\ y_{max} = \frac{v_{0}^2 }{2g}[/tex]

Now we can substitute the value of ymax

[tex]v_{1}^{2}  = v_{0}^{2}  - 2g(\frac{v_{0}^2 }{4g})\\ v_{1}^{2}  = \frac{v_{0}^2 }{2}\\ v_{1} = \frac{v_{0} }{\sqrt{2} }[/tex]

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)

[tex]v=v_{0}-gt\\ t = \frac{v_{0}}{g}[/tex]

This gives us total time from bottom to top

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

[tex]t_{1}  = \frac{v_{1}}{g}\\ t_{1} = \frac{v_{0} }{g\sqrt{2} }[/tex]

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

[tex]t - t_{1}  = \frac{v_{0} }{g} (1 - \frac{1}{\sqrt{2}})[/tex]

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

[tex]\frac{t_{1}}{t - t_{1}}  = 2.4[/tex]