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
a 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 atteint par la balle lors de son 5e rebond. (arrondis ta réponse

a millième près)

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

rebondir?

Explanation:

Part 1: Determine the exponential function rule

Step 1: Define the exponential model

Let the height after \( n \) rebounds be \( h(n) = a \cdot r^n \), where \( a \) is the initial height (before any rebounds, but here the first drop is 1.5m, and after the first rebound, the height is \( 1.5r \), second rebound \( 1.5r^2 \), third rebound \( 1.5r^3 \)). We know at \( n = 3 \), \( h(3) = 0.324 \) m and \( a = 1.5 \) m.

Step 2: Solve for the common ratio \( r \)

Substitute into the formula: \( 0.324 = 1.5 \cdot r^3 \)
Divide both sides by 1.5: \( r^3 = \frac{0.324}{1.5} = 0.216 \)
Take the cube root: \( r = \sqrt[3]{0.216} = 0.6 \)

Step 3: Write the function rule

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

Step 1: Identify \( n \) for the 5th rebound

For the 5th rebound, \( n = 5 \).

Step 2: Substitute into the function

Use \( h(5) = 1.5 \cdot (0.6)^5 \)
Calculate \( (0.6)^5 = 0.6 \times 0.6 \times 0.6 \times 0.6 \times 0.6 = 0.07776 \)
Then \( h(5) = 1.5 \times 0.07776 = 0.11664 \)

Step 3: Round to the thousandth

Rounded to the thousandth, \( 0.117 \) (since the fourth decimal is 4, which is less than 5, but wait, 0.11664 rounded to the thousandth: the third decimal is 6, fourth is 6, so we round up the third decimal: 0.117? Wait, 0.11664: the thousandth place is 6 (0.116...), the next digit is 6, so we round up the 6 to 7? Wait, 0.11664: positions are tenths (1), hundredths (1), thousandths (6), ten - thousandths (6). So when rounding to the thousandth, we look at the ten - thousandth digit (6), which is ≥5, so we add 1 to the thousandth digit: 6 + 1 = 7. So 0.117.

Wait, actually, \( 1.5\times0.07776 = 0.11664 \), which rounds to 0.117 when rounded to the thousandth (three decimal places).

Step 1: Understand the stopping condition

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 can consider when \( h(n) \leq 0 \), but since \( 1.5\cdot(0.6)^n>0 \) for all real \( n \), we consider when the height is so small that it's effectively 0 (e.g., less than a very small positive number like \( \epsilon \), but in the context of the problem, we can think about when the height is 0, but since the function is exponential and never actually reaches 0, we can consider when the height is less than a practical minimum, like 0 (theoretically, it never stops, but maybe in the context of the problem, we can see when \( h(n) = 0 \), but since \( 1.5\cdot(0.6)^n = 0 \) has no solution (exponential functions with positive base and coefficient never reach 0). However, maybe there is a mis - interpretation: maybe the initial drop is 1.5m (before any rebounds), and the first rebound is \( n = 1 \), second \( n = 2 \), etc. Wait, maybe the model is that after \( n \) rebounds, the height is \( h(n) \). But since \( (0.6)^n \) approaches 0 as \( n \) approaches infinity, the ball never actually stops rebounding in the mathematical model. But maybe the problem considers when the height is less than a certain small value, like 0 (but that's not possible) or maybe there is a miscalculation. Wait, maybe the initial height is 1.5m (drop), then after 1st rebound: \( 1.5\times0.6 \), 2nd: \( 1.5\times0.6^2 \), 3rd: \( 1.5\times0.6^3 = 0.324 \) (which matches). So the function is correct. Since \( (0.6)^n>0 \) for all \( n\geq0 \), the ball never stops rebounding in the exponential model. But maybe the problem expects us to consider when the height is 0, which is never, or maybe there is a mistake in the problem. Alternatively, maybe we consider when \( h(n)\leq0 \), but since \( h(n)=1.5\times(0.6)^n>0 \) for all \( n \), there is no finite number of rebounds after which the ball stops. But this seems odd. Maybe the problem has a different interpretation. Wait, maybe the "rebound" is counted as after each hit, so the first rebound is \( n = 1 \), height \( 1.5\times0.6 \), second \( n = 2 \), \( 1.5\times0.6^2 \), third \( n = 3 \), \( 1.5\times0.6^3 = 0.324 \). Since the function is always positive, the ball never stops rebounding in the mathematical sense. But maybe the problem expects an answer like "never" or "infinite number of rebounds".

Answer:

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

Part 2: Height at the 5th rebound