Question

If a ball is thrown into the air at 64 feet per second from the top of a 112​-foot-tall ​building, its height can be modeled by the function Sequals112plus64tminus16tsquared​, where S is in feet and t is in seconds.

If a ball is thrown into the air at 64 feet per second from the top of a 112​-foot-tall ​building, its height can be modeled by the function Sequals112plus64tminus16tsquared​, where S is in feet and t is in seconds.

Answer 1

The height \( S \) of the ball at time \( t \) seconds is modeled by the function:

\[ S(t) = 112 + 64t - 16t^2 \]

Where:
- \( S(t) \) is the height of the ball in feet,
- \( t \) is the time in seconds,
- The ball starts 112 feet above the ground,
- It has an initial velocity of 64 feet per second,
- The term \( -16t^2 \) represents the effect of gravity on the ball, causing it to decelerate.

### Key Points of the Function

1. **Finding the Height at Specific Times:**
- You can substitute different values of \( t \) into the equation to find out how high the ball is at that specific time.

2. **Maximum Height:**
- To find the time at which the ball reaches its maximum height, you can use the vertex formula for a quadratic function. The time at which it reaches its maximum height \( t \) can be found using:
\[
t = -\frac{b}{2a}
\]
where \( a = -16 \) and \( b = 64 \):
\[
t = -\frac{64}{2 \cdot -16} = 2 \text{ seconds}
\]
Then, substitute \( t = 2 \) back into the height function to find the maximum height:
\[
S(2) = 112 + 64(2) - 16(2^2)
\]
\[
S(2) = 112 + 128 - 64 = 176 \text{ feet}
\]
So the maximum height of the ball is 176 feet.

3. **Finding When the Ball Hits the Ground:**
- To find when the ball hits the ground, set \( S(t) = 0 \):
\[
0 = 112 + 64t - 16t^2
\]
Rearranging gives:
\[
16t^2 - 64t - 112 = 0
\]
Dividing the entire equation by 16:
\[
t^2 - 4t - 7 = 0
\]
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1}
\]
\[
t = \frac{4 \pm \sqrt{16 + 28}}{2} = \frac{4 \pm \sqrt{44}}{2}
\]
\[
t = \frac{4 \pm 2\sqrt{11}}{2} = 2 \pm \sqrt{11}
\]
Since time cannot be negative, we take the positive value:
\[
t = 2 + \sqrt{11} \approx 5.32 \text{ seconds}
\]

### Summary
- The maximum height of the ball is 176 feet, reached at 2 seconds.
- The ball hits the ground approximately 5.32 seconds after being thrown.