Let's solve each part step-by-step:

Part 4: Calculating Vertical Jump Height

For this part, we're dealing with the downward portion of the motion, which is essentially the free fall from the maximum height. The data provided is:

• $a = -9.8 \, \text{m/s}^2$ (due to gravity acting downward)
• $v_0 = 0 \, \text{m/s}$ (at the peak of the jump, velocity is zero)
• $t = \text{hang time} / 2$

The equation we’ll use to find the displacement (vertical jump height, $y$) is:

$y = v_0 \cdot t + \frac{1}{2} a t^2$

Since $v_0 = 0$, this simplifies to:

$y = \frac{1}{2} a t^2$

Step 1: Calculate $t$ for each trial

The average hang time was calculated as $0.733$ seconds. Therefore, the downward time portion is:

$t = \frac{0.733}{2} = 0.3665 \, \text{seconds}$

Step 2: Substitute values into the equation

$y = \frac{1}{2} \cdot (-9.8) \cdot (0.3665)^2$

$y = -4.9 \cdot (0.3665)^2$

$y \approx -4.9 \cdot 0.1343$

$y \approx -0.658 \, \text{m}$

Since height is a distance, the negative sign indicates the direction (downward), so the vertical height is approximately $0.658 \, \text{m}$.

Part 5: Calculating Initial Velocity

We need to find the initial velocity at the beginning of the jump using the following values:

• $y = 0.658 \, \text{m}$
• $a = -9.8 \, \text{m/s}^2$
• $v = 0 \, \text{m/s}$ at the peak of the jump
• $t$ is the upward portion of the hang time, which is $0.3665 \, \text{s}$

We can use the following kinematic equation:

$v^2 = v_0^2 + 2ay$

Since $v = 0$ at the peak, we rearrange to find $v_0$:

$0 = v_0^2 + 2(-9.8)(0.658)$

$v_0^2 = 2 \cdot 9.8 \cdot 0.658$

$v_0^2 \approx 12.884$

$v_0 \approx \sqrt{12.884}$

$v_0 \approx 3.59 \, \text{m/s}$

So, the initial velocity at the beginning of the jump is approximately $3.59 \, \text{m/s}$.

Summary:

1. Height of the jump: $0.658 \, \text{m}$
2. Initial velocity: $3.59 \, \text{m/s}$

Would you like further details on any part? Here are some related questions:

1. How can I use the entire hang time to verify the calculations?
2. How would the height change if the average hang time was different?
3. Why is the initial velocity found using upward motion when the height is calculated using downward motion?
4. What if the acceleration due to gravity was different in another location? How would that affect the jump?
5. Can the same formulas be applied for a jump on the moon?

Tip: Whenever you calculate vertical motion, breaking the entire motion into upward and downward portions helps to simplify calculations.