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To solve this problem, we need to determine the **take-off angle** of a long-jump athlete based on the provided parameters. The key data provided is:

- Horizontal distance of the jump, \(d_x = 7.73 \, \text{m}\),
- Initial vertical height of the center of mass, \( y_0 = 0.99 \, \text{m}\),
- Maximum vertical height reached, \( y_{\text{max}} = 1.73 \, \text{m} \),
- Final vertical height at landing, \( y_f = 0.12 \, \text{m} \).

We can approach this using the kinematic equations and the principles of projectile motion.

### Step 1: Vertical motion to determine initial vertical velocity
The maximum height \(y_{\text{max}} = 1.73 \, \text{m}\) is achieved when the vertical velocity is zero, so we can use the following kinematic equation to find the vertical component of the initial velocity (\(v_{y0}\)):

\[
v_y^2 = v_{y0}^2 - 2g (y_{\text{max}} - y_0)
\]
At the peak, \(v_y = 0\), so:
\[
0 = v_{y0}^2 - 2g (y_{\text{max}} - y_0)
\]
Solving for \(v_{y0}\):
\[
v_{y0} = \sqrt{2g(y_{\text{max}} - y_0)}
\]
where \(g = 9.81 \, \text{m/s}^2\) is the acceleration due to gravity.

### Step 2: Time of flight calculation
We will now find the total time of flight, knowing the vertical displacement from the initial height (\(y_0 = 0.99 \, \text{m}\)) to the final height (\(y_f = 0.12 \, \text{m}\)):

We use the following kinematic equation for vertical displacement:
\[
y_f = y_0 + v_{y0}t - \frac{1}{2} g t^2
\]
This is a quadratic equation in \(t\) (the time of flight), which we can solve using the quadratic formula.

### Step 3: Horizontal motion to determine take-off angle
Once we know the time of flight \(t\), we use the horizontal distance traveled to find the horizontal component of the initial velocity, \(v_{x0}\). The horizontal motion is uniform, so:
\[
d_x = v_{x0} t
\]
Solving for \(v_{x0}\):
\[
v_{x0} = \frac{d_x}{t}
\]

### Step 4: Determining the take-off angle
Finally, the take-off angle \(\theta\) can be found using the components of the velocity:
\[
\tan(\theta) = \frac{v_{y0}}{v_{x0}}
\]
So, the take-off angle is:
\[
\theta = \arctan\left(\frac{v_{y0}}{v_{x0}}\right)
\]

Let's now calculate this step by step.The take-off angle of the long jump athlete is approximately **25.35 degrees**.

Would you like further explanation on any of the steps, or do you have any other questions?

Here are 5 related questions to consider:
1. How would the take-off angle change if the horizontal distance was increased?
2. How does gravity affect both the vertical and horizontal components of motion?
3. What would happen to the time of flight if the take-off height was lower than the landing height?
4. How can we calculate the optimal angle for a long jump to maximize distance?
5. What role does air resistance play in altering the calculated results in real-world scenarios?

**Tip:** When solving projectile motion problems, always separate the vertical and horizontal motions to simplify the calculations!

A long jump athlete makes jumps a horizontal distance of 7.73 m. His body center of mass started at a vertical height of 0.99 m, reached a maximum height of 1.73 m, and finished at 0.12 m. Determine his take-off angle (i.e., angle of the velocity vector at take-off; degrees; do NOT enter degree unit) at time of take-off.

Learn how to calculate the take-off angle for a long jump athlete, using principles of projectile motion and kinematic equations. The problem involves determining vertical and horizontal velocity components based on given heights and distances.

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