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\begin{tabular}{l}7 A bouncy rubber ball dropped from any height will lose \( 25 \% \) on its nergni on \\ each successive bounce. If dropped from a height of 30 m , the first bounce will \\ have a height of \( 22,5 \mathrm{~m} \) and the second bounce a height of \( 16,875 \mathrm{~m} \). Determine \\ the total vertical distance travelled by the ball from the time it is dropped from \\ a height of 40 m , until it finally comes to rest. Note that the height from which \\ the ball is dropped is 40 m , and not 30 m . \\ \hline\end{tabular}

Ask by Donnelly Elliott. in South Africa
Jan 28,2025

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The total vertical distance traveled by the ball is 280 meters.

Solución

To determine the total vertical distance traveled by the ball dropped from a height of \( 40 \) meters until it comes to rest, we'll analyze each phase of its motion. ### Understanding the Problem 1. **Initial Drop:** - The ball is initially dropped from a height of \( 40 \) meters. 2. **Successive Bounces:** - On each bounce, the ball retains \( 75\% \) of its previous height (since it loses \( 25\% \) of its energy). - This means the height after each bounce is multiplied by \( 0.75 \). 3. **Total Distance Calculation:** - The ball travels **downward** on the initial drop. - For each subsequent bounce, it travels **upward** and then **downward**. ### Step-by-Step Calculation 1. **Initial Drop:** \[ \text{Distance} = 40 \, \text{m} \] 2. **First Bounce:** \[ \text{Height after first bounce} = 40 \times 0.75 = 30 \, \text{m} \] \[ \text{Distance} = 30 \, \text{m (up)} + 30 \, \text{m (down)} = 60 \, \text{m} \] 3. **Second Bounce:** \[ \text{Height after second bounce} = 30 \times 0.75 = 22.5 \, \text{m} \] \[ \text{Distance} = 22.5 \, \text{m (up)} + 22.5 \, \text{m (down)} = 45 \, \text{m} \] 4. **Continuing Indefinitely:** - This pattern continues indefinitely, with each bounce contributing \( 2 \times \text{current height} \) to the total distance. - The heights form a geometric series: \[ 40 + 2(30 + 22.5 + 16.875 + \dots) \] 5. **Summing the Series:** - The heights form an infinite geometric series with: \[ \text{First term} (a) = 30 \, \text{m} \] \[ \text{Common ratio} (r) = 0.75 \] - The sum \( S \) of an infinite geometric series is: \[ S = \frac{a}{1 - r} = \frac{30}{1 - 0.75} = \frac{30}{0.25} = 120 \, \text{m} \] - Since each bounce contributes twice the height (up and down), the total distance from bounces is: \[ 2 \times 120 = 240 \, \text{m} \] 6. **Total Vertical Distance:** \[ \text{Initial drop} + \text{Bounces} = 40 + 240 = 280 \, \text{m} \] ### Conclusion The **total vertical distance** traveled by the ball from the time it is dropped until it comes to rest is: \[ \boxed{280 \text{ meters}} \]

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When the ball is dropped from a height of 40 m, it will bounce according to the formula for decreasing height by 25% with each bounce. The heights of the bounces can be calculated as follows: the first bounce reaches \( 40 \times (1 - 0.25) = 30 \) m, the next reaches \( 30 \times (1 - 0.25) = 22.5 \) m, and this pattern continues indefinitely, creating a geometric sequence. To find the total vertical distance traveled, consider that the ball travels downward and then upward for each bounce. The total distance includes the initial drop of 40 m, the ascent to each subsequent height, and the descent back down. So, the total distance can be calculated as: 1. Initial Drop: 40 m 2. Bounces: - First bounce: 30 m up + 30 m down - Second bounce: 22.5 m up + 22.5 m down - Continue until the height becomes negligible. Using the formula for the sum of an infinite geometric series where the first term is \( a = 30 \) and the common ratio \( r = 0.75 \), we can find the total distance traveled by the bouncing ball as: \[ \text{Total Distance} = 40 + 2 \times \left( \frac{a}{1 - r} \right) = 40 + 2 \times \left( \frac{30}{1 - 0.75} \right) = 40 + 2 \times 120 = 40 + 240 = 280 \text{ m} \] So, the total vertical distance traveled by the ball until it comes to rest is **280 m**.

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