Question
A golf ball is hit with a club from the ground on a flat surface. The height of the ball, h of t, in feet, t seconds after it was hit can be modeled by the function h of t equals 40 t minus 16 t squared.
Move numbers to the blanks to describe the path of the ball.
Response area with 4 blank spaces
When the ball is hit at 0 seconds, it has a height of
Blank space 1 empty
feet. The ball's height increases until it reaches its maximum height of
Blank space 2 empty
feet after
Blank space 3 empty
seconds. The ball's height then decreases until it reaches the ground
Blank space 4 empty
seconds after it was hit.
Answer options with 9 options.
Move numbers to the blanks to describe the path of the ball.
Response area with 4 blank spaces
When the ball is hit at 0 seconds, it has a height of
Blank space 1 empty
feet. The ball's height increases until it reaches its maximum height of
Blank space 2 empty
feet after
Blank space 3 empty
seconds. The ball's height then decreases until it reaches the ground
Blank space 4 empty
seconds after it was hit.
Answer options with 9 options.
Answers
GPT-4o mini
To determine the values to fill in the blanks, we can analyze the function \( h(t) = 40t - 16t^2 \).
1. When the ball is hit at \( t = 0 \):
\[ h(0) = 40(0) - 16(0)^2 = 0 \quad \text{(height of the ball at 0 seconds)} \]
**Blank 1**: 0
2. To find the maximum height, we need to find the vertex of the parabola represented by the function. The formula for the time at which the maximum height occurs in a quadratic function \( h(t) = at^2 + bt + c \) is given by \( t = -\frac{b}{2a} \). Here, \( a = -16 \) and \( b = 40 \):
\[
t = -\frac{40}{2 \times -16} = \frac{40}{32} = 1.25 \quad \text{(time at which the max height occurs)}
\]
To find the maximum height, we substitute \( t = 1.25 \) back into the height equation:
\[
h(1.25) = 40(1.25) - 16(1.25)^2 = 50 - 16 \cdot 1.5625 = 50 - 25 = 25 \quad \text{(maximum height)}
\]
**Blank 2**: 25
3. The ball reaches the ground when the height \( h(t) = 0 \).
\[
0 = 40t - 16t^2
\]
Factoring gives:
\[
0 = t(40 - 16t)
\]
This yields \( t = 0 \) (when it was hit) and \( 40 - 16t = 0 \) leads to:
\[
16t = 40 \right) t = \frac{40}{16} = 2.5 \quad \text{(time at which the ball hits the ground)}
\]
**Blank 4**: 2.5
So, here are the filled blanks:
- When the ball is hit at 0 seconds, it has a height of **0** feet. The ball's height increases until it reaches its maximum height of **25** feet after **1.25** seconds. The ball's height then decreases until it reaches the ground **2.5** seconds after it was hit.
1. When the ball is hit at \( t = 0 \):
\[ h(0) = 40(0) - 16(0)^2 = 0 \quad \text{(height of the ball at 0 seconds)} \]
**Blank 1**: 0
2. To find the maximum height, we need to find the vertex of the parabola represented by the function. The formula for the time at which the maximum height occurs in a quadratic function \( h(t) = at^2 + bt + c \) is given by \( t = -\frac{b}{2a} \). Here, \( a = -16 \) and \( b = 40 \):
\[
t = -\frac{40}{2 \times -16} = \frac{40}{32} = 1.25 \quad \text{(time at which the max height occurs)}
\]
To find the maximum height, we substitute \( t = 1.25 \) back into the height equation:
\[
h(1.25) = 40(1.25) - 16(1.25)^2 = 50 - 16 \cdot 1.5625 = 50 - 25 = 25 \quad \text{(maximum height)}
\]
**Blank 2**: 25
3. The ball reaches the ground when the height \( h(t) = 0 \).
\[
0 = 40t - 16t^2
\]
Factoring gives:
\[
0 = t(40 - 16t)
\]
This yields \( t = 0 \) (when it was hit) and \( 40 - 16t = 0 \) leads to:
\[
16t = 40 \right) t = \frac{40}{16} = 2.5 \quad \text{(time at which the ball hits the ground)}
\]
**Blank 4**: 2.5
So, here are the filled blanks:
- When the ball is hit at 0 seconds, it has a height of **0** feet. The ball's height increases until it reaches its maximum height of **25** feet after **1.25** seconds. The ball's height then decreases until it reaches the ground **2.5** seconds after it was hit.