Pregunta
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(I) A ball is dropped from a height of 10 m and bounces continuously. With each successive bounce, the ball reaches a height that is \( 50 \% \) of the previous height. If this motion continues indefinitely, what is the total vertical distance travelled by the ball over its entire joumey? (g) The diameter of the first circle in an infinite sequence of circles is 12 cm . Each subscquent circle's diameter is two thirds of the diameter of the previous circle. Calculate (1) the total cireumference of all the circles. (2) the total area of the all the circles.

Ask by Wells Higgins. in South Africa
Feb 03,2025

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- **Problem (I):** The ball travels a total of **30 meters** vertically. - **Problem (g):** - **(1) Total Circumference:** \( 36\pi \) cm. - **(2) Total Area:** \( \frac{324}{5}\pi \) cm².

Solución

Let's tackle each problem step by step. --- ### **Problem (I): Total Vertical Distance Traveled by the Ball** **Given:** - Initial height, \( h_0 = 10 \) meters. - Each bounce reaches \( 50\% \) of the previous height. **Approach:** 1. **Initial Drop:** - The ball first drops \( h_0 = 10 \) meters. 2. **Subsequent Bounces:** - After the first bounce, it rises to \( h_1 = 0.5 \times h_0 = 5 \) meters and then drops the same distance. - This pattern continues indefinitely, with each bounce reaching half the height of the previous one. 3. **Total Distance Calculation:** - **First Drop:** \( 10 \) meters. - **Bounces:** Each bounce contributes \( 2 \times h_n \) meters (up and down). - The heights form a geometric series: \( 5 + 2.5 + 1.25 + \ldots \) - **Sum of Infinite Geometric Series:** \[ \text{Sum} = \frac{a}{1 - r} = \frac{5}{1 - 0.5} = 10 \text{ meters} \] - **Total Distance:** \[ 10 + 2 \times 10 = 30 \text{ meters} \] **Answer:** The total vertical distance traveled by the ball is **30 meters**. --- ### **Problem (g): Infinite Sequence of Circles** **Given:** - Diameter of the first circle, \( d_1 = 12 \) cm. - Each subsequent circle's diameter is \( \frac{2}{3} \) of the previous diameter. **Approach:** #### **(1) Total Circumference of All Circles** **Circumference of a Circle:** \[ C = \pi d \] **Sequence of Diameters:** \[ d_n = 12 \left(\frac{2}{3}\right)^{n-1} \text{ cm} \] **Sequence of Circumferences:** \[ C_n = \pi d_n = 12\pi \left(\frac{2}{3}\right)^{n-1} \text{ cm} \] **Sum of Infinite Geometric Series:** \[ \text{Total Circumference} = \sum_{n=1}^{\infty} C_n = 12\pi \sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n = 12\pi \times \frac{1}{1 - \frac{2}{3}} = 36\pi \text{ cm} \] #### **(2) Total Area of All Circles** **Area of a Circle:** \[ A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \] **Sequence of Areas:** \[ A_n = \frac{\pi d_n^2}{4} = \frac{\pi (12)^2}{4} \left(\frac{2}{3}\right)^{2(n-1)} = 36\pi \left(\frac{4}{9}\right)^{n-1} \text{ cm}^2 \] **Sum of Infinite Geometric Series:** \[ \text{Total Area} = \sum_{n=1}^{\infty} A_n = 36\pi \sum_{n=0}^{\infty} \left(\frac{4}{9}\right)^n = 36\pi \times \frac{1}{1 - \frac{4}{9}} = 36\pi \times \frac{9}{5} = \frac{324}{5}\pi \text{ cm}^2 \] **Answers:** 1. **Total Circumference:** \( 36\pi \) centimeters. 2. **Total Area:** \( \frac{324}{5}\pi \) square centimeters. --- **Summary of Answers:** - **Problem (I):** The total vertical distance traveled by the ball is **30 meters**. - **Problem (g):** - **(1) Total Circumference:** \( 36\pi \) cm. - **(2) Total Area:** \( \frac{324}{5}\pi \) cm².

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When a ball is dropped from a height of 10 m and bounces back to 50% of its previous height, it creates a fascinating pattern. The total vertical distance can be calculated by considering both the drop and the bounces. The total distance is 10 m (drop) + 5 m (first bounce) + 2.5 m + 1.25 m + ... This forms a geometric series where the sum is \( 10 + 2 \times \frac{10}{2} = 10 + 10 = 20 \) m. So, the ball travels a total vertical distance of 20 meters! Now, let's jump into the circle conundrum! The first circle has a diameter of 12 cm, giving it a circumference of \( \pi \times 12 \) cm. Each subsequent circle has a diameter that is two-thirds of the previous one, meaning the second circle has a diameter of 8 cm, the third 5.33 cm, and so on! This creates another geometric series for circumference: \( 12\pi + 8\pi + ... \) This sums to \( \frac{12\pi}{1 - \frac{2}{3}} = 36\pi \) cm for total circumference. For the areas, using \( A = \pi \left(\frac{d}{2}\right)^2 \) leads to the areas summing-up nicely to \( \frac{36\pi}{1 - \frac{2}{3}} = 108 \pi \) cm²!

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