The question provides the equation for the distance $s$ (in feet) of the ball from the ground at any time $t$ (in seconds):

$s = 48t - 16t^2$

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

The ball strikes the ground when $s = 0$. Setting the equation to 0 and solving for $t$:

$48t - 16t^2 = 0$

Factor the equation:

$t(48 - 16t) = 0$

This gives two solutions:

$t = 0 \quad \text{or} \quad 48 - 16t = 0$

Solving for $t$ in the second equation:

$48 = 16t \implies t = \frac{48}{16} = 3$

So, the ball strikes the ground at $t = 3$ seconds.

Part (b): For what time is the ball more than 32 feet above the ground?

We need to solve for when $s > 32$. Set up the inequality:

$48t - 16t^2 > 32$

Rearranging the inequality:

$48t - 16t^2 - 32 > 0 \implies -16t^2 + 48t - 32 > 0$

Divide the entire inequality by $-16$ (and reverse the inequality sign):

$t^2 - 3t + 2 < 0$

Factor the quadratic:

$(t - 1)(t - 2) < 0$

Using a sign chart or analyzing the inequality, the solution is $1 < t < 2$. Therefore, the ball is more than 32 feet above the ground for $1 < t < 2$ seconds.

Final Answers:

• (a) The ball strikes the ground at $t = 3$ seconds.
• (b) The ball is more than 32 feet above the ground for $1 < t < 2$.

Would you like further clarification on any step?

Here are 5 related questions to explore:

1. How would air resistance affect this calculation?
2. What is the maximum height the ball reaches?
3. How would the initial velocity affect the time it takes to hit the ground?
4. Could you solve the same problem using a graphical method?
5. What happens if the ball was thrown downward instead of upward?

Tip: When factoring inequalities, always pay attention to the sign changes across the intervals to determine where the solution holds.