The problem asks to determine the velocity of the object when it impacts the ground. The velocity of the object at any time $t = a$ is given by the limit:

$\lim_{t \to a} \frac{s(a) - s(t)}{a - t}$

This expression is the definition of the derivative of the position function $s(t)$ at $t = a$, which represents the object's velocity function $v(t) = s'(t)$.

Step 1: Find the velocity function

The position function is given as:

$s(t) = -4.9t^2 + 200$

We can differentiate $s(t)$ to find the velocity function $v(t)$:

$v(t) = \frac{d}{dt} [-4.9t^2 + 200] = -9.8t$

Step 2: Determine when the object hits the ground

The object hits the ground when its height $s(t) = 0$. Set the position function equal to zero and solve for $t$:

$-4.9t^2 + 200 = 0$ $4.9t^2 = 200$ $t^2 = \frac{200}{4.9} \approx 40.82$ $t \approx \sqrt{40.82} \approx 6.39 \text{ seconds}$

Step 3: Calculate the velocity at impact

Now, substitute $t = 6.39$ into the velocity function:

$v(6.39) = -9.8 \times 6.39 \approx -62.62 \text{ m/s}$

The negative sign indicates that the object is falling downward. Thus, the velocity at impact is approximately:

$\boxed{-62.62 \text{ m/s}}$

Conclusion:

The object will impact the ground at a velocity of approximately $-62.62$ m/s. The negative sign indicates the direction of the velocity, which is downward.

Would you like more details or have any questions?

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Here are 5 related questions to explore:

1. How would the velocity change if the object started at a different height?
2. How would air resistance affect the velocity at impact?
3. What is the object's average velocity during its fall?
4. What is the acceleration due to gravity in this problem, and how is it derived?
5. Can you find the time when the object's velocity is half its final velocity?

Tip: Differentiation of position functions gives velocity, while the second derivative gives acceleration.