A ball is thrown vertically upwards from the ground at a speed of 40 m/s. The ball then hits the ground. The total distance traveled during the ball's bounce is ….. (e = 0.5 ; g = 10 m/s2)

To calculate the total distance traveled by the ball during its motion and bounce, we break the motion into two parts: the upward and downward journeys. After the ball hits the ground, it bounces up with a reduced velocity due to the coefficient of restitution \( e \). Here’s the step-by-step approach:

### Step 1: Motion during the first throw (up and down)
The initial speed of the ball is \( u = 40 \, \text{m/s} \), and acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \). The ball travels upwards, and at the highest point, its velocity becomes zero.

Using the formula:
\[
v^2 = u^2 - 2gh \quad \text{(where \( v = 0 \) at the highest point)}
\]
\[
0 = 40^2 - 2(10)h_1
\]
\[
h_1 = \frac{40^2}{2 \times 10} = 80 \, \text{m}
\]
So, the ball reaches a maximum height of \( h_1 = 80 \, \text{m} \).

Since the distance covered during the upward journey is 80 m, the ball will also travel 80 m downward to hit the ground. Therefore, the total distance during the first upward and downward journey is:
\[
\text{Distance in first journey} = 80 \, \text{m} \, \text{(up)} + 80 \, \text{m} \, \text{(down)} = 160 \, \text{m}
\]

### Step 2: Motion during the bounce (up and down)
When the ball hits the ground, its velocity is reduced by the coefficient of restitution \( e = 0.5 \).

The velocity just before the bounce is the same as the initial velocity but in the opposite direction, so it’s \( 40 \, \text{m/s} \) downward. The velocity after the bounce becomes:
\[
v_{\text{after}} = e \times v_{\text{before}} = 0.5 \times 40 = 20 \, \text{m/s}
\]

Now, with an initial velocity of \( 20 \, \text{m/s} \), we calculate the maximum height reached after the bounce using the same formula:
\[
v^2 = u^2 - 2gh_2
\]
\[
0 = 20^2 - 2(10)h_2
\]
\[
h_2 = \frac{20^2}{2 \times 10} = 20 \, \text{m}
\]

So, the ball reaches a height of \( h_2 = 20 \, \text{m} \).

The total distance during the bounce (up and down) is:
\[
\text{Distance in bounce} = 20 \, \text{m} \, \text{(up)} + 20 \, \text{m} \, \text{(down)} = 40 \, \text{m}
\]

### Step 3: Further bounces (optional)
We could continue this process for further bounces, where the velocity keeps decreasing by the factor of \( e = 0.5 \) each time. However, if we are only asked about the first bounce, then the total distance traveled is:

### Total Distance
\[
\text{Total Distance} = 160 \, \text{m} + 40 \, \text{m} = 200 \, \text{m}
\]

Thus, the total distance traveled by the ball during its motion and bounce is \( \boxed{200 \, \text{m}} \).

---

Would you like more details or have any questions?

Here are some related problems:
1. How would the total distance change if the coefficient of restitution were 0.8?
2. What is the maximum height if the initial velocity was 50 m/s?
3. How long does it take for the ball to hit the ground after the first throw?
4. What is the velocity of the ball after the second bounce?
5. How many bounces will occur before the ball comes to rest?

**Tip**: In problems involving vertical motion, the coefficient of restitution helps calculate how much energy is retained after each bounce, which reduces both the velocity and height each time.

A ball is thrown vertically upwards from the ground at a speed of 40 m/s. The ball then hits the ground. The total distance traveled during the ball's bounce is ….. (e = 0.5 ; g = 10 m/s^2)

Learn how to calculate the total distance traveled by a ball thrown vertically upwards with an initial speed of 40 m/s and a coefficient of restitution of 0.5. This problem involves kinematics and the law of restitution, providing step-by-step solutions suitable for high school physics students in Grades 9-12.

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