To solve this problem, we need to analyze the motion of the tennis ball in two dimensions: horizontal (x) and vertical (y). We will apply the standard equations of projectile motion to find the necessary details.

Given Data:

• Initial height of the ball: $h_1$ (we will keep this as a variable).
• Horizontal distance to the net: $X_1 = 12.0 \, \text{m}$.
• Height of the net: $h_2 = 1.00 \, \text{m}$.
• Angle of projection: $\theta = 10^\circ$.
• Time when the ball hits the net: $t = 0.400 \, \text{seconds}$.

Goal:

We need to find the initial velocity $v_0$ required for the ball to hit the top of the net after 0.400 seconds. We can divide the problem into two parts: horizontal motion and vertical motion.

1. Horizontal Motion:

In horizontal motion, the velocity remains constant because there is no horizontal acceleration (ignoring air resistance). The horizontal component of the initial velocity is:

$v_{0x} = v_0 \cos(\theta)$

The horizontal distance traveled is given by:

$X_1 = v_{0x} t = v_0 \cos(\theta) \cdot t$

We know $X_1 = 12.0 \, \text{m}$ and $t = 0.400 \, \text{seconds}$, so:

$12.0 = v_0 \cos(10^\circ) \cdot 0.400$

Solving for $v_0$:

$v_0 = \frac{12.0}{\cos(10^\circ) \cdot 0.400}$

2. Vertical Motion:

For vertical motion, the acceleration due to gravity is $g = 9.81 \, \text{m/s}^2$. The vertical component of the velocity is:

$v_{0y} = v_0 \sin(\theta)$

The vertical displacement is:

$\Delta y = h_2 - h_1$

Using the equation for vertical displacement under constant acceleration:

$\Delta y = v_{0y} t - \frac{1}{2} g t^2$

Substituting values:

$1.00 - h_1 = v_0 \sin(10^\circ) \cdot 0.400 - \frac{1}{2} \cdot 9.81 \cdot (0.400)^2$

Solving the System of Equations:

We now have two equations:

1. $v_0 = \frac{12.0}{\cos(10^\circ) \cdot 0.400}$
2. $1.00 - h_1 = v_0 \sin(10^\circ) \cdot 0.400 - \frac{1}{2} \cdot 9.81 \cdot (0.400)^2$

We can solve these equations to find $v_0$ and then substitute it into the second equation to find $h_1$. Let me calculate this for you.The initial velocity $v_0$ required for Serena Williams to hit the ball so that it reaches the net in 0.400 seconds is approximately 30.46 m/s.

Additionally, the height from which the ball was hit, $h_1$, is found to be approximately 0.33 m below the height of the net, meaning $h_1 \approx 0.67 \, \text{m}$ (since the net's height is 1.00 m).

Would you like further details on the calculations or have any questions? Here are some related questions you could explore:

1. How would the result change if the angle of projection was 15° instead of 10°?
2. How would air resistance affect the motion of the ball?
3. What would happen if the net height was 1.5 m instead of 1.0 m?
4. How does increasing the initial velocity affect the time to hit the net?
5. What is the maximum height the ball reaches during its flight?

Tip: Understanding the breakdown of vertical and horizontal motion is key to solving projectile motion problems.