We are given the equation for the height of the ball at time $t$ as:

$s(t) = -16t^2 - 48t + 87$

We need to determine when the height $s(t) = 59$ feet. Set $s(t) = 59$:

$-16t^2 - 48t + 87 = 59$

Now, solve for $t$:

1. Subtract 59 from both sides:

$-16t^2 - 48t + 87 - 59 = 0$

$-16t^2 - 48t + 28 = 0$

2. Divide the entire equation by $-4$ to simplify:

$4t^2 + 12t - 7 = 0$

Now solve this quadratic equation using the quadratic formula:

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

Here, $a = 4$, $b = 12$, and $c = -7$. Plugging these into the quadratic formula:

$t = \frac{-12 \pm \sqrt{12^2 - 4(4)(-7)}}{2(4)}$

$t = \frac{-12 \pm \sqrt{144 + 112}}{8}$

$t = \frac{-12 \pm \sqrt{256}}{8}$

$t = \frac{-12 \pm 16}{8}$

Now, solve for both possibilities:

1. $t = \frac{-12 + 16}{8} = \frac{4}{8} = \frac{1}{2}$

2. $t = \frac{-12 - 16}{8} = \frac{-28}{8} = -\frac{7}{2}$

Since time cannot be negative, the only valid solution is:

$t = \frac{1}{2} \text{ seconds}$

Thus, the ball is at a height of 59 feet at $t = \frac{1}{2}$ seconds.

Would you like further details or have any questions?

Here are 5 related questions to expand this concept:

1. What is the velocity of the ball at $t = 0.5$ seconds?
2. At what time does the ball hit the ground?
3. How long does it take for the ball to reach its maximum height?
4. What is the ball's maximum height?
5. How would the equation change if the ball was thrown upward instead?

Tip: Always check the domain of your solution when working with time-related equations in real-world scenarios.