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The problem involves a ball being thrown vertically upward with an initial velocity of 160 ft/s, and the height of the ball after \( t \) seconds is given by the equation:

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

We are tasked with solving two parts of the problem:

### (a) What is the maximum height reached by the ball?
To find the maximum height, we can use the fact that the velocity at the maximum height is zero. The velocity is the derivative of the position function \( s(t) \) with respect to time.

The velocity function \( v(t) \) is:
\[
v(t) = \frac{ds}{dt} = 160 - 32t
\]

Setting \( v(t) = 0 \) to find the time when the ball reaches its maximum height:
\[
160 - 32t = 0
\]
\[
t = \frac{160}{32} = 5 \, \text{seconds}
\]

Now, plug \( t = 5 \) seconds into the position function to find the maximum height:
\[
s(5) = 160(5) - 16(5)^2 = 800 - 16(25) = 800 - 400 = 400 \, \text{feet}
\]

Thus, the maximum height is **400 feet**.

### (b) What is the velocity of the ball when it is 384 ft above the ground on its way up and on its way down?

We need to find the times \( t \) when the ball is at a height of 384 ft. Set \( s(t) = 384 \) in the height equation:
\[
160t - 16t^2 = 384
\]
Rearranging gives:
\[
16t^2 - 160t + 384 = 0
\]
Dividing through by 16:
\[
t^2 - 10t + 24 = 0
\]
Solving this quadratic equation using the quadratic formula:
\[
t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(24)}}{2(1)} = \frac{10 \pm \sqrt{100 - 96}}{2} = \frac{10 \pm \sqrt{4}}{2} = \frac{10 \pm 2}{2}
\]

So, the two possible times are:
\[
t = \frac{10 + 2}{2} = 6 \, \text{seconds} \quad \text{(on the way down)}
\]
\[
t = \frac{10 - 2}{2} = 4 \, \text{seconds} \quad \text{(on the way up)}
\]

Now, use the velocity equation \( v(t) = 160 - 32t \) to find the velocity at these times:

- For \( t = 4 \, \text{seconds} \) (on the way up):
\[
v(4) = 160 - 32(4) = 160 - 128 = 32 \, \text{ft/s}
\]

- For \( t = 6 \, \text{seconds} \) (on the way down):
\[
v(6) = 160 - 32(6) = 160 - 192 = -32 \, \text{ft/s}
\]

Thus, the velocity is **32 ft/s** on the way up and **-32 ft/s** on the way down.

---

Would you like further details on any step? Here are 5 related questions to deepen understanding:

1. What is the significance of a negative velocity in the context of this problem?
2. How would the time to reach maximum height change if the initial velocity were different?
3. Can you derive the general formula for the maximum height of an object thrown upwards with any initial velocity?
4. What are the units of acceleration in this problem, and why is it constant?
5. How would the equation change if the object was thrown from a height above the ground?

Tip: Always check units when solving physics problems to ensure consistency!

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?

This problem involves finding the maximum height and velocities of a ball thrown vertically with an initial velocity of 160 ft/s. Using calculus and kinematic equations, the maximum height is determined, and velocities at a given height of 384 ft are calculated. Suitable for high school students.

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