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.



Hint: Find v0 to reach y_max in terms of g and y_max and recall the velocity at y_max is zero. Then find v1 to reach y_max/2 with the same kinematic equation. The time to reach y_max is obtained from v0=g (t), and the time to reach y_max/2 is given by v1-v0= -g(t1). Now, t1 is the time to reach y_max/2, and the quantity t-t1 is the time to go from y_max/2 to y_max. You want the ratio of (t-t1)/t1

Answer 1

Answer on Question 62782, Physics, Mechanics, Relativity

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 $y_{max}$ be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he pends above $y_{max}/2$ to the time it takes him to go from the floor to that height. You may ignore air resistance.

Solution:

Let's first find the velocity $v_{1}$ of an athlete to reach half of the maximum height from the kinematic equation:

$$v_{1}^{2} = v_{0}^{2} - 2gh = v_{0}^{2} - 2g\left(\frac{y_{max}}{2}\right),$$

here, $v_{0}$ is the initial velocity of an athlete, $v_{1}$ is the velocity of an athlete at half the maximum height, $g$ is the acceleration due to gravity, $h = y_{max}/2$ is the half of the maximum height.

We can find the maximum height that an athlete can reach from the Law of Conservation of Energy:

$$KE = PE,$$ $$\frac{1}{2}mv_{0}^{2} = mgy_{max},$$ $$y_{max} = \frac{v_{0}^{2}}{2g}.$$

Then, substituting $y_{max}$ into the first equation we get:

$$v_{1}^{2} = v_{0}^{2} - 2g\left(\frac{v_{0}^{2}}{4g}\right),$$ $$v_{1}^{2} = v_{0}^{2} - \frac{v_{0}^{2}}{2} = \frac{v_{0}^{2}}{2},$$ $$v _ {1} = \frac {v _ {0}}{\sqrt {2}}.$$

We can find the time $t$ that an athlete needs to reach the maximum height $(y_{\text{max}})$ from the kinematic equation:

$$v = v _ {0} - g t,$$

here, $v$ is the final velocity of an athlete at the maximum height, $v_{0}$ is the initial velocity of an athlete.

Since, $v = 0 \, ms^{-1}$ , we get:

$$t = \frac {v _ {0}}{g}.$$

Similarly, we can find the time $t_1$ that an athlete needs to reach maximum height from the $(y_{\text{max}} / 2)$ :

$$t _ {1} = \frac {v _ {1}}{g} = \frac {v _ {0}}{g \sqrt {2}}$$

So, it is obviously, that the time to reach $y_{max}$ from $y_{max} / 2$ (or the time he is above $y_{max} / 2$ moving up) is nothing more than the difference between $t$ and $t_1$ :

$$t - t _ {1} = \frac {v _ {0}}{g} \left(1 - \frac {1}{\sqrt {2}}\right),$$

Finally, we can calculate the ratio of the time he is above $y_{\text{max}} / 2$ to the time it takes him to go from the floor to that height:

$$\frac {t _ {1}}{t - t _ {1}} = \frac {v _ {0}}{g \sqrt {2}} \frac {g \sqrt {2}}{v _ {0} (\sqrt {2} - 1)} = 2.4$$

**Answer:**

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

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