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The question asks whether the average velocity of a ball dropped from the CN Tower (450 meters above the ground) over a time interval starting at \( t = 0 \) and ending at \( t = 5 \) seconds is given by the following limit expression:

\[
\lim_{h \to 0} \frac{4.9(5+h)^2 - 4.9(5)^2}{h}
\]

### Problem Breakdown:
1. **Distance Function:** 
   The distance fallen by the ball as a function of time \( t \) is given by the equation \( s = f(t) = 4.9t^2 \). This comes from the equation of motion for free-fall under gravity (where \( 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity, and dividing by 2 gives \( 4.9 \)).

2. **Average Velocity Over Time Interval:**
   The average velocity over a time interval \( [0, 5] \) is the change in distance divided by the change in time:
   \[
   \text{Average velocity} = \frac{f(5) - f(0)}{5 - 0}
   \]
   where \( f(t) = 4.9t^2 \).

   Substituting into the above formula:
   \[
   \text{Average velocity} = \frac{4.9(5)^2 - 4.9(0)^2}{5 - 0} = \frac{4.9(25)}{5} = 24.5 \, \text{m/s}
   \]

3. **Limit Expression:**
   The limit provided in the problem,
   \[
   \lim_{h \to 0} \frac{4.9(5+h)^2 - 4.9(5)^2}{h},
   \]
   is a form of the derivative of the function \( f(t) = 4.9t^2 \) at \( t = 5 \). This limit calculates the instantaneous velocity at \( t = 5 \), not the average velocity over the interval. Let’s compute the derivative.

4. **Instantaneous Velocity:**
   The derivative of \( s = 4.9t^2 \) with respect to \( t \) gives the instantaneous velocity:
   \[
   v(t) = \frac{d}{dt}[4.9t^2] = 9.8t.
   \]
   At \( t = 5 \),
   \[
   v(5) = 9.8(5) = 49 \, \text{m/s}.
   \]

### Conclusion:
The limit provided in the question is **not** the average velocity but the instantaneous velocity at \( t = 5 \), which is 49 m/s. The correct average velocity over the interval \( [0, 5] \) is 24.5 m/s.

Thus, the statement is **false**.

Would you like further details or have any questions?

### Related Questions:
1. How is the equation \( s = 4.9t^2 \) derived for a free-falling object?
2. What is the difference between average velocity and instantaneous velocity?
3. How do you compute the derivative of a quadratic function?
4. What is the significance of the limit definition of a derivative?
5. How does gravity affect the velocity of an object in free fall?

**Tip:** The derivative provides the slope of the tangent line to the curve, which represents instantaneous velocity at a given time.

True or false: If a ball is dropped from the upper observation deck of the CN Tower, 450 m above the ground, then the average velocity over the time interval that starts t = 0 and ends at t = 5 is given by lim (h→0) [(4.9(5+h)^2 - 4.9(5)^2) / h]. You can assume the distance the object has fallen is given by s = f(t) = 4.9t^2. Justify your answer.

Explore whether the average velocity of a ball dropped from the CN Tower over a 5-second interval matches the limit expression provided, using calculus concepts. Analyze the difference between average and instantaneous velocity through derivatives. Suitable for students studying introductory calculus.

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