use the position function s(t) = −4.9t^2 + 360, which gives the height (in meters) of an object that has fallen for t seconds from a height of 360 meters. The velocity at time t = a seconds is given by lim t→a(s(a) − s(t))/(a − t)
When will the object hit the ground? At what velocity, v, will the object impact the ground?

### 1. **Finding the time when the object hits the ground**:

The object will hit the ground when its height \( s(t) \) becomes zero. We can find this by setting the position function \( s(t) \) equal to zero and solving for \( t \).

The position function is:
\[
s(t) = -4.9t^2 + 360
\]
We want to find \( t \) such that \( s(t) = 0 \):
\[
0 = -4.9t^2 + 360
\]
Rearranging this equation:
\[
4.9t^2 = 360
\]
\[
t^2 = \frac{360}{4.9}
\]
\[
t^2 = 73.47
\]
Taking the square root of both sides:
\[
t = \sqrt{73.47} \approx 8.57 \, \text{seconds}
\]
So, the object will hit the ground after approximately **8.57 seconds**.

### 2. **Finding the velocity at the time of impact**:

The velocity function is the derivative of the position function \( s(t) \). To find the velocity at any time \( t \), we differentiate \( s(t) \):

\[
v(t) = \frac{d}{dt} \left( -4.9t^2 + 360 \right) = -9.8t
\]

Now, substitute \( t = 8.57 \) seconds (the time when the object hits the ground) into the velocity equation:

\[
v(8.57) = -9.8 \times 8.57
\]
\[
v(8.57) \approx -83.99 \, \text{m/s}
\]

Thus, the object will impact the ground at a velocity of approximately **-83.99 meters per second** (the negative sign indicates downward motion).

### Summary:
- The object hits the ground at **8.57 seconds**.
- The velocity upon impact is approximately **-83.99 m/s**.

Would you like further details or have any questions about this solution?

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Here are 5 related questions:
1. How does the velocity of the object change over time?
2. What is the average velocity during the fall?
3. How would the results change if the object started from a different height?
4. What is the significance of the acceleration term in the position function?
5. Could we calculate the object's velocity at halfway through its fall?

**Tip**: The derivative of the position function gives you the velocity, and the second derivative provides the acceleration!

Use the position function s(t) = −4.9t^2 + 360, which gives the height (in meters) of an object that has fallen for t seconds from a height of 360 meters. The velocity at time t = a seconds is given by lim t→a(s(a) − s(t))/(a − t). When will the object hit the ground? At what velocity, v, will the object impact the ground?

Explore how to determine when an object hits the ground using the quadratic position function s(t) = -4.9t^2 + 360 and calculate the impact velocity using the derivative. This problem is ideal for high school or college students studying kinematics and calculus.

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