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 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, the initial height \( h_0 = 1.5 \) m (the height before the first rebound, but we know the height at the 3rd rebound is 0.324 m). So we have \( h(3) = 1.5 \cdot r^3 = 0.324 \).

Step 2: Solve for \( r \)

We solve the equation \( 1.5 \cdot r^3 = 0.324 \) for \( r \). First, divide both sides by 1.5: \( r^3 = \frac{0.324}{1.5} = 0.216 \). Then take the cube root of both sides: \( r = \sqrt[3]{0.216} = 0.6 \).

Step 3: Write the function rule

Now that we have \( r = 0.6 \) and \( h_0 = 1.5 \), 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 use the function \( h(n) = 1.5 \cdot (0.6)^n \) with \( n = 5 \).

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

First, calculate \( 0.6^5 \). We know that \( 0.6^3 = 0.216 \), \( 0.6^4 = 0.6^3 \cdot 0.6 = 0.216 \cdot 0.6 = 0.1296 \), and \( 0.6^5 = 0.6^4 \cdot 0.6 = 0.1296 \cdot 0.6 = 0.07776 \).

Step 3: Multiply by the initial height

Now multiply by 1.5: \( h(5) = 1.5 \cdot 0.07776 = 0.11664 \).

Step 4: Round to the thousandth

Rounding 0.11664 to the thousandth place (three decimal places) gives 0.117 (since the fourth decimal is 6, which is greater than 5, we round up the third decimal: 0.11664 ≈ 0.117).

Step 1: Understand when the ball stops

The ball stops rebounding when the height \( h(n) \) becomes 0 (or very close to 0, practically when the height is negligible, but mathematically, we solve \( h(n) = 1.5 \cdot (0.6)^n = 0 \). However, an exponential function with a positive base and coefficient never actually reaches 0, but we can consider when the height is so small that it's effectively 0 (e.g., less than a very small positive number like \( 10^{-10} \) or when \( n \) is large enough that \( (0.6)^n \) is negligible). But practically, we can analyze the limit: as \( n \to \infty \), \( (0.6)^n \to 0 \), so the height approaches 0. However, in reality, we can consider when the height is less than a certain threshold (e.g., 0.001 m, 0.0001 m, etc.). Let's solve for \( n \) when \( h(n) \leq \epsilon \) (a small \( \epsilon \), say \( \epsilon = 0.001 \) m).

Step 2: Solve \( 1.5 \cdot (0.6)^n \leq \epsilon \)

Let's use \( \epsilon = 0.001 \) m. Then:
\( (0.6)^n \leq \frac{0.001}{1.5} \approx 0.0006667 \)
Take the natural logarithm of both sides:
\( n \cdot \ln(0.6) \leq \ln(0.0006667) \)
Since \( \ln(0.6) \) is negative (because 0.6 < 1), when we divide both sides, the inequality sign flips:
\( n \geq \frac{\ln(0.0006667)}{\ln(0.6)} \)
Calculate:
\( \ln(0.0006667) \approx -7.3132 \)
\( \ln(0.6) \approx -0.5108 \)
So \( n \geq \frac{-7.3132}{-0.5108} \approx 14.32 \)
Since \( n \) must be an integer, we take \( n = 15 \) (because at \( n = 14 \), the height is still above \( \epsilon \), and at \( n = 15 \), it's below). Let's verify:
For \( n = 14 \): \( h(14) = 1.5 \cdot (0.6)^{14} \). Calculate \( (0.6)^{14} \approx 0.6^{10} \cdot 0.6^4 \approx (0.0060466) \cdot (0.1296) \approx 0.000783 \), so \( h(14) \approx 1.5 \cdot 0.000783 \approx 0.001175 \) m (still above 0.001 m).
For \( n = 15 \): \( (0.6)^{15} = (0.6)^{14} \cdot 0.6 \approx 0.000783 \cdot 0.6 \approx 0.0004698 \), so \( h(15) \approx 1.5 \cdot 0.0004698 \approx 0.0007047 \) m (below 0.001 m).

However, if we consider the theoretical limit, the exponential function never actually reaches 0, but in practice, after a large number of rebounds (around 15 or more, depending on the threshold), the height is so small that the ball effectively stops rebounding.

Alternatively, if we consider the height approaching 0 as \( n \to \infty \), the ball never truly stops in the mathematical model, but practically, after a finite number of rebounds when the height is negligible.

Answer:

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

Question 2: Find the height reached by the ball on its 5th rebound (round to the thousandth).