Question
A ball is launched upward at 64 feet per second from a platform that is 36 feet high.
The function that models this situation is h(t)=−16t2+64t+36
where t is the time in seconds and h(t)
is the height of the ball.
Find the maximum height the ball reaches and how long it will take to get there.
(1 point)
A) What key feature of the graph is the problem asking you to find?
(answer choices: the vertex, y-intercept, x-intercept)
B) The ball will reach a maximum height of
feet in
seconds.
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The function that models this situation is h(t)=−16t2+64t+36
where t is the time in seconds and h(t)
is the height of the ball.
Find the maximum height the ball reaches and how long it will take to get there.
(1 point)
A) What key feature of the graph is the problem asking you to find?
(answer choices: the vertex, y-intercept, x-intercept)
B) The ball will reach a maximum height of
feet in
seconds.
Skip to navigation
Answers
GPT-4o mini
To find the maximum height of the ball and how long it takes to reach that height, we need to analyze the function \( h(t) = -16t^2 + 64t + 36 \).
**A)** The problem is asking you to find the **vertex** of the parabola represented by this function. The vertex will give us the maximum height because the parabola opens downwards (the coefficient of \( t^2 \) is negative).
**B)** To find the time at which the maximum height occurs, we can use the vertex formula for a quadratic function of the form \( ax^2 + bx + c \):
\[
t = -\frac{b}{2a}
\]
In this equation:
- \( a = -16 \)
- \( b = 64 \)
Plugging in the values:
\[
t = -\frac{64}{2 \times -16} = \frac{64}{32} = 2 \text{ seconds}
\]
Next, we substitute \( t = 2 \) back into the height function \( h(t) \) to find the maximum height:
\[
h(2) = -16(2^2) + 64(2) + 36
\]
\[
h(2) = -16(4) + 128 + 36
\]
\[
h(2) = -64 + 128 + 36
\]
\[
h(2) = 64 + 36 = 100 \text{ feet}
\]
Thus, the answers are:
**B)** The ball will reach a maximum height of **100 feet** in **2 seconds**.
**A)** The problem is asking you to find the **vertex** of the parabola represented by this function. The vertex will give us the maximum height because the parabola opens downwards (the coefficient of \( t^2 \) is negative).
**B)** To find the time at which the maximum height occurs, we can use the vertex formula for a quadratic function of the form \( ax^2 + bx + c \):
\[
t = -\frac{b}{2a}
\]
In this equation:
- \( a = -16 \)
- \( b = 64 \)
Plugging in the values:
\[
t = -\frac{64}{2 \times -16} = \frac{64}{32} = 2 \text{ seconds}
\]
Next, we substitute \( t = 2 \) back into the height function \( h(t) \) to find the maximum height:
\[
h(2) = -16(2^2) + 64(2) + 36
\]
\[
h(2) = -16(4) + 128 + 36
\]
\[
h(2) = -64 + 128 + 36
\]
\[
h(2) = 64 + 36 = 100 \text{ feet}
\]
Thus, the answers are:
**B)** The ball will reach a maximum height of **100 feet** in **2 seconds**.