Let's break this down using physics and kinematics.

Given Data:

• Speed of the serve: $v_0 = 170 \, \text{km/h} = 47.22 \, \text{m/s}$ (convert to meters per second)
• Height of the ball at serving: $h_0 = 2.5 \, \text{m}$
• Angle of the serve: $\theta = 6.120^\circ$ below the horizontal
• Distance from the net to the service line: $d_{\text{service}} = 11.9 \, \text{m}$
• Height of the net: $h_{\text{net}} = 0.91 \, \text{m}$

Step 1: Break down the initial velocity

We decompose the initial velocity into horizontal and vertical components:

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

Step 2: Vertical motion analysis

The ball is under the influence of gravity, so we use the following kinematic equation for vertical motion:

$y = h_0 + v_{0y}t - \frac{1}{2} g t^2$

Where:

• $y$ is the height of the ball at any time $t$,
• $g = 9.81 \, \text{m/s}^2$ is the acceleration due to gravity.

Step 3: Horizontal motion analysis

In horizontal motion, the ball travels at constant velocity, so the horizontal distance is given by:

$x = v_{0x} t$

Step 4: Solving for time when ball crosses the net

To find the time when the ball crosses the net, set the horizontal distance to the distance from the net, $x = 11.9 \, \text{m}$. We'll use this to solve for the time $t$ it takes for the ball to reach the net.

Step 5: Find if ball clears the net

After finding $t$, substitute it into the vertical motion equation to check if $y \geq h_{\text{net}}$ when the ball reaches the net.

Step 6: Distance the ball lands

Finally, use the total time the ball is in the air (when it hits the ground, $y = 0$) to find how far from the service line the ball lands.

I'll calculate this step-by-step for you.### Results:

• Time to reach the net: $0.253 \, \text{seconds}$
• Height of the ball when it crosses the net: $3.46 \, \text{m}$ (well above the net's height of $0.91 \, \text{m}$, so the ball clears the net)
• Distance from the service line where the ball lands: $65.38 \, \text{m}$

This means that the ball lands about 65.38 meters from the service line, indicating it travels quite far, likely beyond the bounds of a normal tennis court.

Let me know if you need further clarification or details!

Here are five related questions you might explore:

1. How would the landing distance change if the angle of serve was increased?
2. What would happen if the initial speed was reduced by 10%?
3. How would the results differ if the net height were raised to 1.2 meters?
4. What is the effect of air resistance on the trajectory of the ball?
5. How long would it take the ball to travel if the court were 30 meters long?

Tip: Remember that air resistance can significantly affect high-speed objects like tennis balls, making the real-life trajectory shorter.