Question

nom :
chapitre 2.4 la fonction exponentielle
problème de compréhension
les démarches complètes sont obligatoires.
attention au plagiat.
on laisse tomber une balle dune hauteur de 1,5 m. chaque fois que la balle touche
aut sol, la balle rebondit. lors de son 3e rebond, elle atteint une hauteur de 0,324 m.

1. détermine la règle de la fonction exponentielle qui modélise cette situation.
2. trouve la hauteur atteinte par la balle lors de son 5e rebond. (arrondis ta réponse

aumillième près)

1. selon le modèle exponentiel, après combien de rebonds la balle cessera-t-elle de

rebondir?

Explanation:

Question 1: Determine the rule of the exponential function modeling this situation.

Step 1: Define the exponential model

An exponential function for the height \( h \) after \( n \) rebounds can be written as \( h(n) = h_0 \cdot r^n \), where \( h_0 \) is the initial height, \( r \) is the rebound ratio, and \( n \) is the number of rebounds. Here, \( h_0 = 1.5 \) m (initial height before any rebound, but actually, the first rebound is after the first drop. Wait, let's check the problem: when \( n = 3 \) (3rd rebound), \( h(3) = 0.324 \) m. So the formula is \( h(n) = 1.5 \cdot r^n \), because after \( n \) rebounds, the height is \( 1.5 \times r^n \).

Step 2: Solve for \( r \)

We know that when \( n = 3 \), \( h(3) = 0.324 \). So:

$$ 0.324 = 1.5 \cdot r^3 $$

Divide both sides by 1.5:

$$ r^3 = \frac{0.324}{1.5} = 0.216 $$

Take the cube root of both sides:

$$ r = \sqrt[3]{0.216} = 0.6 $$

So the exponential function is \( h(n) = 1.5 \cdot (0.6)^n \), where \( n \) is the number of rebounds.

Step 1: Use the exponential function

We have \( h(n) = 1.5 \cdot (0.6)^n \). For \( n = 5 \):

$$ h(5) = 1.5 \cdot (0.6)^5 $$

Step 2: Calculate \( (0.6)^5 \)

First, \( 0.6^2 = 0.36 \), \( 0.6^3 = 0.216 \), \( 0.6^4 = 0.1296 \), \( 0.6^5 = 0.07776 \)

Step 3: Multiply by 1.5

$$ h(5) = 1.5 \times 0.07776 = 0.11664 $$

Round to the thousandth: \( 0.117 \) m (wait, 0.11664 rounded to the thousandth is 0.117? Wait, 0.11664: the third decimal is 6, the next digit is 6, so we round up: 0.117? Wait, no: 0.11664. The thousandth place is the third digit after the decimal: 0.116 (thousandth is 6), the next digit is 6, so we round up the 6 to 7? Wait, 0.11664: 0.1 (tenths), 0.11 (hundredths), 0.116 (thousandths), and the next digit is 6, so we round the thousandth place up: 0.117. Wait, but let's check the calculation again. \( 0.6^5 = 0.07776 \). Then \( 1.5 \times 0.07776 = 0.11664 \). Rounded to the thousandth (three decimal places): look at the fourth decimal, which is 4? Wait, no: 0.11664 is 0.1 (1), 0.11 (1), 0.116 (6), 0.1166 (6), 0.11664 (4). Wait, no, 0.11664: the digits are 1 (tenths), 1 (hundredths), 6 (thousandths), 6 (ten - thousandths), 4 (hundred - thousandths). So to round to the thousandth (three decimal places), we look at the ten - thousandth place, which is 6. Since 6 ≥ 5, we round the thousandth place up: 6 becomes 7. So 0.117.

Step 1: Understand when the ball stops

The ball stops rebounding when the height \( h(n) \) is approximately 0 (or very close to 0, practically, when the height is so small that it's negligible, like approaching 0). So we need to find \( n \) such that \( 1.5 \cdot (0.6)^n \approx 0 \). But mathematically, an exponential function with base \( 0 < r < 1 \) approaches 0 as \( n \) approaches infinity. However, in practice, we can find when the height is less than a very small value, say 0 (theoretically, it never truly stops, but in the model, we can consider when \( h(n) \) is effectively 0, or when the height is less than a tiny positive number, like 0. But since it's a model, maybe we consider when \( h(n) \) becomes 0, but that never happens. Wait, maybe the problem means when the height is less than a certain threshold, like 0 (theoretically, as \( n \to \infty \), \( h(n) \to 0 \)). But maybe in the context of the problem, we can solve for \( n \) when \( h(n) = 0 \), but that's not possible. Alternatively, maybe when the height is less than a very small value, like 0.001 m. Let's try to solve \( 1.5 \cdot (0.6)^n < \epsilon \), where \( \epsilon \) is a small number (e.g., 0). But since \( (0.6)^n \) is always positive, the height is always positive, but approaches 0. So in theory, the ball never stops rebounding in the exponential model, but practically, we can find when the height is very small. Let's check for \( n \) such that \( 1.5 \cdot (0.6)^n \leq 0 \), but that's impossible. Alternatively, maybe the problem expects us to consider when the height is 0, but since the exponential function never reaches 0, we can say that in the exponential model, the ball never stops rebounding (it approaches 0 as \( n \) approaches infinity). But maybe there's a misinterpretation. Wait, maybe the initial height is 1.5 m, and after the first drop, it rebounds. Wait, maybe the formula is \( h(n) = 1.5 \cdot (0.6)^n \), and we need to find when \( h(n) \) is 0. But since \( 1.5>0 \) and \( (0.6)^n>0 \) for all \( n \geq 0 \), \( h(n) \) is never 0. So in the exponential model, the ball never stops rebounding (it continues to rebound with height approaching 0 as the number of rebounds increases without bound).

Answer:

\( h(n) = 1.5 \times (0.6)^n \) (where \( n \) is the number of rebounds)

Question 2: Find the height at the 5th rebound.