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
what is the maxium height
what is the maxium height
Answers
GPT-4o mini
To find the maximum height of the ball modeled by the function:
\[ S(t) = 112 + 64t - 16t^2 \]
we recognize that this is a quadratic function in the standard form \( S(t) = at^2 + bt + c \), where:
- \( a = -16 \)
- \( b = 64 \)
- \( c = 112 \)
The vertex of a parabola given in this form occurs at \( t = -\frac{b}{2a} \).
Substituting the values of \( a \) and \( b \):
\[
t = -\frac{64}{2 \cdot (-16)} = \frac{64}{32} = 2 \text{ seconds}
\]
Now, we substitute \( t = 2 \) back into the height function \( S(t) \) to find the maximum height:
\[
S(2) = 112 + 64(2) - 16(2^2)
\]
\[
= 112 + 128 - 16(4)
\]
\[
= 112 + 128 - 64
\]
\[
= 176 \text{ feet}
\]
Therefore, the maximum height of the ball is **176 feet**.
\[ S(t) = 112 + 64t - 16t^2 \]
we recognize that this is a quadratic function in the standard form \( S(t) = at^2 + bt + c \), where:
- \( a = -16 \)
- \( b = 64 \)
- \( c = 112 \)
The vertex of a parabola given in this form occurs at \( t = -\frac{b}{2a} \).
Substituting the values of \( a \) and \( b \):
\[
t = -\frac{64}{2 \cdot (-16)} = \frac{64}{32} = 2 \text{ seconds}
\]
Now, we substitute \( t = 2 \) back into the height function \( S(t) \) to find the maximum height:
\[
S(2) = 112 + 64(2) - 16(2^2)
\]
\[
= 112 + 128 - 16(4)
\]
\[
= 112 + 128 - 64
\]
\[
= 176 \text{ feet}
\]
Therefore, the maximum height of the ball is **176 feet**.