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The image presents a physics problem involving a baseball pitch modeled by two parametric equations:

\[
x(t) = 144t
\]
\[
y(t) = -16t^2 + 6t + 5
\]

where \( x(t) \) and \( y(t) \) represent the horizontal and vertical positions of the baseball at time \( t \) in seconds, and the units are in feet.

### Given Questions and Their Solutions

1. **(a) Calculate the horizontal velocity of the baseball at time \( t \).**
   - The horizontal velocity \( x'(t) \) is the derivative of \( x(t) \) with respect to \( t \), which yields:
     \[
     x'(t) = 144 \, \text{ft/sec}
     \]

2. **(b) What is the horizontal velocity of the baseball when it passes over home plate?**
   - Since \( x'(t) = 144 \, \text{ft/sec} \) is constant, the horizontal velocity at home plate remains \( 144 \, \text{ft/sec} \).

3. **(c) What is the vertical velocity of the baseball at time \( t \)?**
   - The vertical velocity \( y'(t) \) is the derivative of \( y(t) \) with respect to \( t \):
     \[
     y'(t) = -32t + 6
     \]

4. **(d) Speed of the baseball when it passes over home plate**
   - The speed \( s(t) \) is given by the magnitude of the velocity vector, calculated as:
     \[
     s(t) = \sqrt{[x'(t)]^2 + [y'(t)]^2}
     \]
   - At the time \( t \) when the baseball reaches home plate (60.5 feet), solve \( x(t) = 60.5 \) for \( t \):
     \[
     144t = 60.5 \Rightarrow t = \frac{60.5}{144} \approx 0.4201 \, \text{sec}
     \]
   - Substituting \( t = 0.4201 \) into \( y'(t) = -32(0.4201) + 6 \):
     \[
     y'(t) \approx -7.4432 \, \text{ft/sec}
     \]
   - Calculating \( s(t) \):
     \[
     s(t) = \sqrt{144^2 + (-7.4432)^2} \approx 144.1924 \, \text{ft/sec}
     \]
   - Converting to mph:
     \[
     s(t) \approx 144.1924 \times 0.681818 \approx 98.28 \, \text{mph}
     \]

5. **(e) Time when the baseball hits the ground ( \( y(t) = 0 \))**
   - Solve \( -16t^2 + 6t + 5 = 0 \) for \( t \). Using the quadratic formula:
     \[
     t = \frac{-6 \pm \sqrt{6^2 - 4 \cdot (-16) \cdot 5}}{2 \cdot (-16)}
     \]
     \[
     t = \frac{-6 \pm \sqrt{196}}{-32} = \frac{-6 \pm 14}{-32}
     \]
   - The two solutions are:
     \[
     t = \frac{8}{-32} = 0.5 \, \text{sec} \quad \text{and} \quad t = \frac{-20}{-32} = 1.25 \, \text{sec}
     \]
   - Since \( t = 1.25 \, \text{sec} \) is positive, this is when the ball hits the ground.

### Additional Questions
Would you like further details on any part of the solution? Here are some related questions to extend understanding:

1. What would change in the problem if air resistance were considered?
2. How would different initial velocities affect the trajectory of the baseball?
3. Can we model other sports scenarios using similar parametric equations?
4. How would we determine the angle at which the baseball was thrown?
5. How does this problem relate to projectile motion concepts in physics?

#### Tip:
When solving parametric equations, always analyze both components (horizontal and vertical) independently before combining them to find the total speed or time of flight.

A major league baseball pitcher throws a pitch modeled by parametric equations x(t) = 144t and y(t) = -16t^2 + 6t + 5. Given that the distance between the pitcher and home plate is 60.5 feet, calculate the horizontal velocity, vertical velocity, speed, and the time at which the baseball hits the ground.

Solve a projectile motion problem involving a baseball pitch modeled by parametric equations. Learn to calculate the horizontal and vertical velocity, total speed, and time to ground impact for a baseball thrown over 60.5 feet. This solution covers key calculus concepts and is suitable for high school or early college students.

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