Serving at a speed of 170 km/h, a tennis player hits ball at a height of 2.5 m and angle θ below the horizontal. The service line is 11.9 m from net, which is 0.91 m high. what distance, from the service line does the ball land θ=6.120

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.

Serving at a speed of 170 km/h, a tennis player hits ball at a height of 2.5 m and angle θ below the horizontal. The service line is 11.9 m from net, which is 0.91 m high. What distance from the service line does the ball land θ=6.120?

Learn how to calculate the landing distance of a tennis ball served at 170 km/h and 6.12° below the horizontal. This problem covers kinematic equations, trigonometric functions, and projectile motion. Suitable for high school physics students (Grades 11-12).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation