https://file.aiquickdraw.com/mathbot/file/1729989146548l2yqror0.jpg

To solve this problem, we’ll address both parts:

Given:
\[ s(t) = 64 + 112t - 16t^2 \]
where \( s(t) \) is the height of the ball after \( t \) seconds.

### Part (a): Maximum Height
To find the maximum height, we need to locate the vertex of the quadratic function \( s(t) \), since it opens downwards (the \( -16t^2 \) term is negative).

1. The time \( t \) at which the maximum height is reached can be found using the formula:
   \[
   t = -\frac{b}{2a}
   \]
   where \( a = -16 \) and \( b = 112 \).

   Substituting:
   \[
   t = -\frac{112}{2 \times -16} = \frac{112}{32} = 3.5 \text{ seconds}
   \]

2. Now, substitute \( t = 3.5 \) back into \( s(t) \) to find the maximum height:
   \[
   s(3.5) = 64 + 112(3.5) - 16(3.5)^2
   \]
   Simplifying this:
   \[
   s(3.5) = 64 + 392 - 16 \times 12.25
   \]
   \[
   s(3.5) = 64 + 392 - 196 = 260 \text{ feet}
   \]

Thus, the maximum height the ball reaches is **260 feet**.

### Part (b): Velocity When the Ball Hits the Ground
To find the velocity when the ball hits the ground, we need to:

1. **Determine the time \( t \) when the ball reaches ground level** (\( s(t) = 0 \)):
   \[
   0 = 64 + 112t - 16t^2
   \]
   Rearranging this, we get:
   \[
   16t^2 - 112t - 64 = 0
   \]
   Divide through by 16:
   \[
   t^2 - 7t - 4 = 0
   \]
   Solve this quadratic equation using the quadratic formula:
   \[
   t = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}
   \]
   \[
   t = \frac{7 \pm \sqrt{49 + 16}}{2}
   \]
   \[
   t = \frac{7 \pm \sqrt{65}}{2}
   \]
   Approximating \( \sqrt{65} \approx 8.06 \):
   \[
   t = \frac{7 \pm 8.06}{2}
   \]
   So:
   - \( t \approx \frac{15.06}{2} = 7.53 \) seconds (we take the positive root since time cannot be negative).

2. **Calculate the velocity** when the ball hits the ground using the derivative of \( s(t) \), which represents velocity:
   \[
   v(t) = s'(t) = 112 - 32t
   \]
   Substitute \( t = 7.53 \):
   \[
   v(7.53) = 112 - 32 \times 7.53
   \]
   \[
   v(7.53) = 112 - 240.96 = -128.96 \text{ ft/sec}
   \]

The velocity of the ball when it hits the ground is approximately **-128.96 ft/sec** (negative because it’s downward).

---

Would you like more details on any step?

### Follow-up Questions:
1. What is the initial velocity of the ball?
2. How long does it take the ball to reach its maximum height?
3. How would the maximum height change if the initial velocity were increased?
4. What is the significance of the negative velocity when the ball hits the ground?
5. How would the time to hit the ground change if the building were taller?

### Tip:
When analyzing a projectile motion problem, remember that the maximum height is reached when the vertical velocity is zero.

If a ball is thrown vertically upward from the roof of a 64-foot building with a velocity of 112 ft/sec, its height after t seconds is s(t) = 64 + 112t - 16t^2.

a.) What is the maximum height the ball reaches?
b.) What is the velocity of the ball when it hits the ground (height 0)?

Calculate the maximum height reached by a ball thrown upward from a 64-foot building at 112 ft/sec, and determine its velocity upon impact with the ground. This problem uses quadratic functions and derivatives, covering concepts in calculus and physics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation