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Let's solve problem #98, which involves analyzing the motion of a ball thrown downward from a height of 220 feet with an initial velocity.

### Problem Statement:
A ball is thrown straight down from the top of a 220-foot building with an initial velocity of \(-22\) feet per second.

- **Part (a):** What is its velocity after 3 seconds?
- **Part (b):** What is its velocity after falling 108 feet?

### Given:
- Initial height \( s_0 = 220 \) feet
- Initial velocity \( v_0 = -22 \) feet per second (negative because it’s downward)
- Position function: \( s(t) = -16t^2 + v_0 t + s_0 \)

### Part (a): Velocity after 3 seconds

1. **Velocity function**: The velocity function is the derivative of the position function with respect to time \( t \).
   \[
   v(t) = \frac{ds(t)}{dt} = -32t + v_0
   \]
   Substituting \( v_0 = -22 \) feet per second:
   \[
   v(t) = -32t - 22
   \]

2. **Velocity at \( t = 3 \) seconds**:
   \[
   v(3) = -32(3) - 22 = -96 - 22 = -118 \text{ feet per second}
   \]

### Part (b): Velocity after falling 108 feet

1. **Determine the time \( t \) when the ball has fallen 108 feet**:
   - The ball falls from 220 feet to \( 220 - 108 = 112 \) feet.
   - Set the position function equal to 112 feet and solve for \( t \):
     \[
     s(t) = -16t^2 - 22t + 220 = 112
     \]
     \[
     -16t^2 - 22t + 108 = 0
     \]

     Solve the quadratic equation:
     \[
     16t^2 + 22t - 108 = 0
     \]
     Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 16 \), \( b = 22 \), and \( c = -108 \):
     \[
     t = \frac{-22 \pm \sqrt{22^2 + 4 \times 16 \times 108}}{2 \times 16}
     \]
     \[
     t = \frac{-22 \pm \sqrt{484 + 6912}}{32}
     \]
     \[
     t = \frac{-22 \pm \sqrt{7396}}{32}
     \]
     \[
     t = \frac{-22 \pm 

Explore the analysis of a ball thrown downward from a 220-foot building with an initial velocity. Learn about its velocity after 3 seconds and after falling 108 feet. This problem covers kinematics and quadratic equations, suitable for high school students.

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