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The problem presented in the image involves a ball being thrown vertically upward with an initial velocity of 64 feet per second, and the equation for the ball's height \( s \) (in feet) after \( t \) seconds is given by:

\[
s = 64t - 16t^2
\]

### Part (a): When will the ball strike the ground?

To find when the ball strikes the ground, we set the height \( s = 0 \) because the ball will be on the ground at that point. This gives the equation:

\[
64t - 16t^2 = 0
\]

We can factor this equation:

\[
16t(4 - t) = 0
\]

This gives two solutions:

\[
t = 0 \quad \text{or} \quad t = 4
\]

Since \( t = 0 \) is when the ball is initially thrown, the ball strikes the ground at \( t = 4 \) seconds.

### Part (b): For what time is the ball more than 28 feet above the ground?

To find when the ball is more than 28 feet above the ground, we solve the inequality:

\[
64t - 16t^2 > 28
\]

First, rearrange the inequality:

\[
64t - 16t^2 - 28 > 0
\]

Divide the entire inequality by 4 to simplify:

\[
16t - 4t^2 - 7 > 0
\]

Rearrange it as a standard quadratic inequality:

\[
-4t^2 + 16t - 7 > 0
\]

Solve this inequality by finding the roots of the equation \( -4t^2 + 16t - 7 = 0 \) using the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = -4 \), \( b = 16 \), and \( c = -7 \). Substituting these values:

\[
t = \frac{-16 \pm \sqrt{16^2 - 4(-4)(-7)}}{2(-4)}
\]
\[
t = \frac{-16 \pm \sqrt{256 - 112}}{-8}
\]
\[
t = \frac{-16 \pm \sqrt{144}}{-8}
\]
\[
t = \frac{-16 \pm 12}{-8}
\]

This gives two solutions:

\[
t = \frac{-16 + 12}{-8} = \frac{-4}{-8} = \frac{1}{2}
\]
\[
t = \frac{-16 - 12}{-8} = \frac{-28}{-8} = 3.5
\]

Thus, the ball is more than 28 feet above the ground between \( t = 0.5 \) seconds and \( t = 3.5 \) seconds.

---

Would you like further explanation on any part of this solution?

Here are 5 related questions:
1. How can we verify the ball's maximum height using calculus?
2. What if the initial velocity was different—how would that change the equations?
3. How long would it take for the ball to reach its highest point?
4. How would this problem change if we were working with meters instead of feet?
5. Can we model air resistance in this scenario and how would it affect the motion?

**Tip:** In motion problems, always start by identifying when the object hits the ground or passes a specific point by setting the height equation to that value.

A ball is thrown vertically upward with an initial velocity of 64 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s = 64t - 16t^2. (a) At what time will the ball strike the ground? (b) For what time t is the ball more than 28 feet above the ground?

This problem involves solving a quadratic equation to determine when a ball, thrown upward with a velocity of 64 feet per second, hits the ground and stays above 28 feet. Learn the step-by-step solution, which includes the use of the quadratic formula.

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