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The problem involves the motion of a ball thrown vertically upward, and its height after \( t \) seconds is given by the equation:

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

Here’s how we solve the parts of the problem:

### Part (a) Maximum Height

To find the maximum height, we need to determine when the velocity of the ball becomes zero (since the ball momentarily stops at its highest point). Velocity is the derivative of the height function:

\[
v(t) = \frac{ds}{dt} = 160 - 32t
\]

Set the velocity equal to zero to find the time when the ball reaches its maximum height:

\[
160 - 32t = 0 \implies t = \frac{160}{32} = 5 \text{ seconds}
\]

Now, substitute \( t = 5 \) into the height equation to find the maximum height:

\[
s(5) = 160(5) - 16(5^2) = 800 - 16(25) = 800 - 400 = 400 \text{ feet}
\]

Thus, the maximum height reached by the ball is **400 feet**.

### Part (b) Velocity when the ball is 384 feet above the ground

We are asked to find the velocity when the ball is at a height of 384 feet both on the way up and on the way down.

First, substitute \( s(t) = 384 \) into the height equation:

\[
160t - 16t^2 = 384
\]

Rearrange this to form a quadratic equation:

\[
16t^2 - 160t + 384 = 0
\]

Divide the whole equation by 16:

\[
t^2 - 10t + 24 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

\[
t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(24)}}{2(1)} = \frac{10 \pm \sqrt{100 - 96}}{2} = \frac{10 \pm \sqrt{4}}{2} = \frac{10 \pm 2}{2}
\]

Thus, we get two solutions:

\[
t = \frac{10 + 2}{2} = 6 \quad \text{and} \quad t = \frac{10 - 2}{2} = 4
\]

So, the ball is at 384 feet at \( t = 4 \) seconds (on the way up) and \( t = 6 \) seconds (on the way down).

#### Velocity at \( t = 4 \) seconds (on the way up):

Use the velocity equation:

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

#### Velocity at \( t = 6 \) seconds (on the way down):

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

The negative sign indicates that the ball is moving downward.

### Summary of Answers:
- **Maximum height:** 400 feet.
- **Velocity at 384 feet (on the way up):** 32 ft/s.
- **Velocity at 384 feet (on the way down):** -32 ft/s.

Would you like more details or have any questions? 

### Here are 5 related questions to expand your understanding:
1. How would the maximum height change if the initial velocity were different?
2. What is the total time the ball stays in the air before hitting the ground?
3. How can we derive the time at which the ball reaches half of its maximum height?
4. What would be the velocity of the ball when it is at half the maximum height?
5. Can you determine the ball's acceleration at any given time?

### Tip:
When solving motion problems involving vertical motion under gravity, remember that the acceleration due to gravity is constant and is usually represented by \(-32 \, \text{ft/s}^2\) in such equations.

If a ball is thrown vertically upward with an initial velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. (Consider up to be the positive direction.)
(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?
What is the velocity (in ft/s) of the ball when it is 384 ft above the ground on its way down?

Learn how to find the maximum height and velocity of a ball thrown vertically with an initial velocity of 160 ft/s. This problem covers concepts from kinematics, quadratic equations, and calculus. Step-by-step solutions provided for finding maximum height and velocity at specific heights. Suitable for high school students in Grades 9-12.

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