A basketball star covers 2.95 m horizontally in a jump to dunk the ball. His motion through space can be modeled precisely as that of a particle at his center of mass. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.80 m above the floor and is at elevation 0.890 m when he touches down again.
(a) Determine his time of flight (his "hang time").

To determine the time of flight, we will treat the basketball star's jump as a projectile motion problem. The center of mass follows a parabolic trajectory, and we will use kinematic equations to analyze the vertical motion.

### Given dataTo determine the time of flight, we will treat the basketball star's jump as a projectile motion problem. The center of mass follows a parabolic trajectory, and we will use kinematic equations to analyze the vertical motion.

### Given 
- Initial height: \( y_0 = 1.02 \, \text{m} \)
- Maximum height: \( y_{\text{max}} = 1.80 \, \text{m} \)
- Final height: \( y_f = 0.890 \, \text{m} \)
- Horizontal distance: \( x = 2.95 \, \text{m} \)

We assume that:
- Gravity \( g = 9.81 \, \text{m/s}^2 \)
- The motion is symmetric, i.e., time to reach maximum height is equal to the time to descend from maximum height.

### Step 1: Time to reach the maximum height
The motion in the vertical direction is governed by the kinematic equation:
\[
v_f^2 = v_0^2 - 2g (y_{\text{max}} - y_0)
\]
At the maximum height, the final vertical velocity \( v_f = 0 \). Rearranging for the initial vertical velocity \( v_0 \):
\[
0 = v_0^2 - 2g (y_{\text{max}} - y_0)
\]
\[
v_0^2 = 2g (y_{\text{max}} - y_0)
\]
\[
v_0 = \sqrt{2g (y_{\text{max}} - y_0)}
\]
Substitute the values:
\[
v_0 = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times (1.80 \, \text{m} - 1.02 \, \text{m})}
\]
\[
v_0 = \sqrt{2 \times 9.81 \times 0.78}
\]
\[
v_0 = \sqrt{15.2756} \approx 3.91 \, \text{m/s}
\]

### Step 2: Total time of flight
Now, we'll use the kinematic equation for the vertical displacement between the initial and final heights:
\[
y_f = y_0 + v_0 t - \frac{1}{2} g t^2
\]
Substitute the known values:
\[
0.890 = 1.02 + 3.91 t - \frac{1}{2} \times 9.81 \, t^2
\]
\[
0 = -0.13 + 3.91 t - 4.905 t^2
\]
Rearrange the equation:
\[
4.905 t^2 - 3.91 t - 0.13 = 0
\]

This is a quadratic equation, which we can solve for \( t \) using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 4.905 \), \( b = -3.91 \), and \( c = -0.13 \).

Now, let's calculate the time.The two solutions for time are:

- \( t_1 = 0.829 \, \text{seconds} \)
- \( t_2 = -0.032 \, \text{seconds} \) (This is not physically meaningful since time cannot be negative.)

Thus, the time to land is \( t = 0.829 \, \text{seconds} \).

### Total time of flight (hang time):
Since the projectile motion is symmetric, the total time of flight is twice the time to reach the maximum height. Therefore, the total time of flight is:
\[
\text{Total time} = 2 \times t_1 = 2 \times 0.829 = 1.658 \, \text{seconds}
\]

The basketball star's hang time is approximately **1.66 seconds**.

Would you like to explore the details or ask any other questions? Here are 5 related problems:

1. How high would the player jump if the horizontal distance was increased?
2. What is the player's initial horizontal velocity?
3. How does the horizontal velocity change during the flight?
4. What would the time of flight be if the maximum height reached was different?
5. How does increasing gravity affect the time of flight?

**Tip:** In projectile motion, the horizontal and vertical components of motion are independent.

A basketball star covers 2.95 m horizontally in a jump to dunk the ball. His motion through space can be modeled precisely as that of a particle at his center of mass. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.80 m above the floor and is at elevation 0.890 m when he touches down again. (a) Determine his time of flight (his 'hang time').

Determine the hang time of a basketball star using projectile motion principles. This problem involves analyzing vertical motion, solving quadratic equations, and applying kinematic formulas. The jump covers 2.95 m horizontally with a maximum height of 1.80 m and an initial height of 1.02 m. Suitable for high school students in Grades 10-12.

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