A ball is thrown downward with a speed of 48ft/s from the top of a building that is 87 feet tall. After t seconds, its height above the ground in feet is given by s(t)=−16t^2−48t+87. When is the ball at a height of 59ft? Enter an exact answer.

We are given the equation for the height of the ball:

\[
s(t) = -16t^2 - 48t + 87
\]

We need to determine the time \( t \) when the ball is at a height of 59 feet, meaning we set \( s(t) = 59 \) and solve for \( t \):

\[
59 = -16t^2 - 48t + 87
\]

Now, subtract 59 from both sides:

\[
0 = -16t^2 - 48t + 87 - 59
\]

\[
0 = -16t^2 - 48t + 28
\]

This is a quadratic equation:

\[
-16t^2 - 48t + 28 = 0
\]

We can simplify this equation by dividing through by \(-4\) to make the coefficients smaller:

\[
4t^2 + 12t - 7 = 0
\]

Now we will solve this quadratic equation using the quadratic formula. The quadratic formula is:

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

For the equation \( 4t^2 + 12t - 7 = 0 \), the coefficients are:
- \( a = 4 \)
- \( b = 12 \)
- \( c = -7 \)

Substitute these values 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 possible values of \( t \):

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, we discard \( t = -\frac{7}{2} \) and keep \( t = \frac{1}{2} \).

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

Would you like any further details or clarification? 

Here are 5 related questions:
1. What is the speed of the ball at \( t = \frac{1}{2} \) seconds?
2. At what time does the ball hit the ground?
3. What is the velocity equation of the ball as a function of time?
4. How long does it take for the ball to reach its maximum velocity?
5. Can you derive the initial height using the given function?

**Tip:** When solving quadratic equations, always check if simplifying the coefficients first makes solving easier.

Learn how to solve the quadratic equation s(t) = -16t^2 - 48t + 87 to find the time when a ball thrown from a building reaches a height of 59 feet. The solution involves applying the quadratic formula. Suitable for students in Grades 9-12.

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