Question

If a ball is thrown upward at 64 feet per second from a height of 12 ​feet, the height of the ball can be modeled by Sequals12plus64tminus16t squared ​feet, where t is the number of seconds after the ball is thrown. How long after the ball is thrown is the height 72 ​feet?
Question content area bottom
Part 1
It takes
  
enter your response here seconds for the ball to reach the height 72 feet.
​(Type an integer or a simplified fraction. Use a comma to separate answers as​ needed.)

Answer 1

To find out how long it takes for the ball to reach a height of 72 feet, we can set the height equation equal to 72 and solve for \( t \).

The height equation is given as:

\[
S = 12 + 64t - 16t^2
\]

We want to find \( t \) when \( S = 72 \):

\[
72 = 12 + 64t - 16t^2
\]

Let's rearrange this equation:

\[
72 - 12 = 64t - 16t^2
\]

\[
60 = 64t - 16t^2
\]

Now let's move everything to one side of the equation:

\[
-16t^2 + 64t - 60 = 0
\]

To simplify this equation, we can multiply through by \(-1\):

\[
16t^2 - 64t + 60 = 0
\]

Next, we can divide the entire equation by 4 to further simplify:

\[
4t^2 - 16t + 15 = 0
\]

Now we can use the quadratic formula to solve for \( t \):

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

where \( a = 4 \), \( b = -16 \), and \( c = 15 \). Plugging in these values:

\[
t = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \cdot 4 \cdot 15}}{2 \cdot 4}
\]

\[
t = \frac{16 \pm \sqrt{256 - 240}}{8}
\]

\[
t = \frac{16 \pm \sqrt{16}}{8}
\]

\[
t = \frac{16 \pm 4}{8}
\]

This gives us two possible solutions:

1. \( t = \frac{20}{8} = \frac{5}{2} = 2.5 \)
2. \( t = \frac{12}{8} = \frac{3}{2} = 1.5 \)

Thus, the times at which the ball reaches a height of 72 feet are:

- \( t = 1.5 \) seconds (when the ball is rising)
- \( t = 2.5 \) seconds (when the ball is falling)

So in answer to your question, it takes **1.5 seconds** and **2.5 seconds** for the ball to reach the height of 72 feet.

You can enter your response as:

**1.5, 2.5**