If a ball is thrown vertically upward with an initial velocity of 160 ft/s, then its height after *t* seconds is 

*s* = 160*t* − 16*t*2.

 If a ball is thrown vertically upward with an initial velocity of 160 ft/s, then its height after *t* seconds is 

*s* = 160*t* − 16*t*2(Consider up to be the positive direction.)

(a)

What is the maximum height (in ft) reached by the ball?

  ft

(b)

What is the velocity (in ft/s) of the ball when it is 384 ft above the ground on its way up?

  ft/s

What is the velocity (in ft/s) of the ball when it is 384 ft above the ground on its way down?

  ft/s

To solve these problems, let's start by analyzing the given equation for the height \( s \) of the ball after \( t \) seconds:

\[
s = 160t - 16t^2
\]

### Part (a): Maximum Height Reached by the Ball

The height \( s \) will be maximized when the velocity \( v = \frac{ds}{dt} \) is zero (since at the maximum height, the ball momentarily stops before descending).

1. **Calculate the derivative of \( s \):**
   \[
   v = \frac{ds}{dt} = 160 - 32t
   \]

2. **Set the velocity \( v \) to zero and solve for \( t \):**
   \[
   160 - 32t = 0
   \]
   \[
   32t = 160
   \]
   \[
   t = 5 \text{ seconds}
   \]

3. **Calculate the maximum height by substituting \( t = 5 \) back into the height equation:**
   \[
   s = 160(5) - 16(5)^2
   \]
   \[
   s = 800 - 16 \cdot 25
   \]
   \[
   s = 800 - 400
   \]
   \[
   s = 400 \text{ ft}
   \]

**Answer for (a):** The maximum height reached by the ball is **400 ft**.

---

### Part (b): Velocity of the Ball at \( s = 384 \) ft

To find the velocity when the ball is at 384 ft, we need to determine the time(s) \( t \) at which \( s = 384 \) ft.

1. **Set up the equation:**
   \[
   384 = 160t - 16t^2
   \]

2. **Rewrite this as a standard quadratic equation:**
   \[
   16t^2 - 160t + 384 = 0
   \]
   \[
   t^2 - 10t + 24 = 0
   \]

3. **Solve this quadratic equation using the quadratic formula:**
   \[
   t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 24}}{2 \cdot 1}
   \]
   \[
   t = \frac{10 \pm \sqrt{100 - 96}}{2}
   \]
   \[
   t = \frac{10 \pm \sqrt{4}}{2}
   \]
   \[
   t = \frac{10 \pm 2}{2}
   \]

   So, the two values of \( t \) are:
   \[
   t = \frac{12}{2} = 6 \text{ seconds (on the way up)}
   \]
   \[
   t = \frac{8}{2} = 4 \text{ seconds (on the way down)}
   \]

4. **Calculate the velocity at \( t = 4 \) and \( t = 6 \) seconds:**

   Using \( v = 160 - 32t \):

   - At \( t = 4 \) seconds:
     \[
     v = 160 - 32(4) = 160 - 128 = 32 \text{ ft/s}
     \]

   - At \( t = 6 \) seconds:
     \[
     v = 160 - 32(6) = 160 - 192 = -32 \text{ ft/s}
     \]

**Answer for (b):**
- The velocity when the ball is 384 ft above the ground on the way up is **32 ft/s**.
- The velocity when the ball is 384 ft above the ground on the way down is **-32 ft/s**.

---

Would you like further details on any of these steps, or is there anything specific you would like to explore further?

---

1. What is the initial velocity and how does it affect the ball's trajectory?
2. How do we derive the time at which the maximum height is reached?
3. Why does the velocity become zero at the highest point of the motion?
4. What are the two types of motion phases in projectile problems like this one?
5. Can we determine the total time the ball is in the air?

**Tip:** When analyzing vertical motion, remember that the acceleration due to gravity will always act downward, even if the initial velocity is upward.

If a ball is thrown vertically upward with an initial velocity of 160 ft/s, then its height after *t* seconds is *s* = 160*t* − 16*t*^2. (a) What is the maximum height (in ft) reached by the ball? (b) What is the velocity (in ft/s) of the ball when it is 384 ft above the ground on its way up and on its way down?

Explore how to calculate the maximum height and velocity of a ball thrown upward with an initial velocity of 160 ft/s using calculus and quadratic equations. Detailed step-by-step solutions for finding the height and velocity at 384 ft above the ground. Suitable for Grades 11-12.

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