4. A rubber ball is dropped from a height of 9 meters and it bounces up to 2
3
of its previous
height after each bounce.
(a) Find the height it will rise after two bounces. 
(b) Calculate the total distance travelled by the ball when it hits the ground for the 10th
time. 
(c) What is the total distance travelled by the ball if it continues to bounce in the same way
indefinitely.

Let's solve the problem step by step:

### Part (a): Height after two bounces

The ball is initially dropped from a height of 9 meters, and after each bounce, it reaches \(\frac{2}{3}\) of its previous height.

- **First bounce**: After the first bounce, the ball rises to \(\frac{2}{3}\) of 9 meters:
  \[
  h_1 = 9 \times \frac{2}{3} = 6 \, \text{meters}
  \]
  
- **Second bounce**: After the second bounce, the ball rises to \(\frac{2}{3}\) of 6 meters:
  \[
  h_2 = 6 \times \frac{2}{3} = 4 \, \text{meters}
  \]

Thus, the height after two bounces is **4 meters**.

---

### Part (b): Total distance travelled by the ball after 10 hits

The ball travels a distance each time it falls and then rises back. For the first drop, it only falls. From then on, it falls and rises alternately.

The total distance travelled by the ball can be computed as:
- The ball first falls 9 meters.
- It rises to \(6\) meters (after the first bounce) and falls back the same distance.
- It then rises to \(4\) meters and falls back to the ground, and so on.

We need to calculate the total distance after 10 hits (i.e., 5 complete bounces).

The pattern of distances is:
- First drop: \(9\) meters.
- First bounce (fall and rise): \(6 + 6 = 12\) meters.
- Second bounce (fall and rise): \(4 + 4 = 8\) meters.
- Third bounce (fall and rise): \( \frac{8}{3} + \frac{8}{3} = \frac{16}{3} \) meters.
- Continue this pattern for 10 hits.

This forms a geometric series. The total distance \(D_{\text{total}}\) after 10 hits is:

\[
D_{\text{total}} = 9 + 2 \times \left(6 + 4 + \frac{8}{3} + \dots\right) 
\]

Let’s calculate this geometric series.

---

### Part (c): Total distance if the ball bounces indefinitely

For an infinite series, we can sum up the total distance travelled by the ball. The total distance is the sum of the distances travelled in both the falling and rising parts.

The total distance travelled indefinitely is:

\[
D = 9 + 2 \times \sum_{n=0}^{\infty} \left( 9 \times \left(\frac{2}{3}\right)^n \right)
\]

This infinite geometric series can be summed using the formula:

\[
S_{\infty} = \frac{a}{1 - r}
\]

Where \(a\) is the first term and \(r\) is the common ratio. We can apply this to find the total distance travelled.

---

Would you like further details on any specific part of this solution? 

Here are some related questions:
1. How do geometric series work?
2. What is the physical meaning of an infinite series in this context?
3. How does air resistance affect real-world bouncing ball problems?
4. Can we model other types of bounce behavior with different ratios?
5. How do changes in initial height affect the total distance travelled?

**Tip:** Understanding geometric progressions is key in many physics and engineering problems related to repeated events, such as bouncing or decay processes.

A rubber ball is dropped from a height of 9 meters and it bounces up to 2/3 of its previous height after each bounce. (a) Find the height it will rise after two bounces. (b) Calculate the total distance travelled by the ball when it hits the ground for the 10th time. (c) What is the total distance travelled by the ball if it continues to bounce in the same way indefinitely.

Explore how to solve a bouncing ball problem using geometric series. Learn to calculate the height after two bounces, the total distance travelled after 10 hits, and the total distance travelled if the ball bounces indefinitely. This problem is suitable for students in Grades 9-11 and covers key concepts such as geometric progressions and ratios.

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