Question
The height y (in feet) of a ball thrown by a child is y=-1/16x^2+4x+3 where x is the horizontal distance in feet from the point at which the ball is thrown.
(a) How high is the ball when it leaves the child's hand?
feet
(b) What is the maximum height of the ball?
feet
(c) How far from the child does the ball strike the ground? Round your answers to the nearest 0.01.
feet
(a) How high is the ball when it leaves the child's hand?
feet
(b) What is the maximum height of the ball?
feet
(c) How far from the child does the ball strike the ground? Round your answers to the nearest 0.01.
feet
Answers
GPT-4o mini
To analyze the function \( y = -\frac{1}{16}x^2 + 4x + 3 \), which describes the height of the ball in terms of the horizontal distance \( x \):
### (a) Height when the ball leaves the child's hand
When the ball leaves the child's hand, \( x = 0 \).
Substituting \( x = 0 \) into the height equation:
\[
y(0) = -\frac{1}{16}(0)^2 + 4(0) + 3 = 3
\]
**So, the height when the ball leaves the child's hand is 3 feet.**
### (b) Maximum height of the ball
The maximum height of a parabolic equation in the form \( y = ax^2 + bx + c \) occurs at the vertex, which can be found using the formula:
\[
x = -\frac{b}{2a}
\]
Here, \( a = -\frac{1}{16} \) and \( b = 4 \).
Calculating the x-coordinate of the vertex:
\[
x = -\frac{4}{2 \cdot -\frac{1}{16}} = -\frac{4}{-\frac{1}{8}} = 4 \cdot 8 = 32
\]
Now substituting \( x = 32 \) back into the height equation to find the maximum height:
\[
y(32) = -\frac{1}{16}(32)^2 + 4(32) + 3
\]
Calculating each term:
\[
y(32) = -\frac{1}{16}(1024) + 128 + 3 = -64 + 128 + 3 = 67
\]
**Thus, the maximum height of the ball is 67 feet.**
### (c) How far from the child does the ball strike the ground?
To find out where the ball strikes the ground, we set \( y = 0 \) and solve for \( x \):
\[
0 = -\frac{1}{16}x^2 + 4x + 3
\]
Multiplying through by -16 to eliminate fractions:
\[
0 = x^2 - 64x - 48
\]
Now we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -64 \), and \( c = -48 \).
Calculating the discriminant:
\[
b^2 - 4ac = (-64)^2 - 4(1)(-48) = 4096 + 192 = 4288
\]
Now applying the quadratic formula:
\[
x = \frac{64 \pm \sqrt{4288}}{2}
\]
Calculating \( \sqrt{4288} \):
\[
\sqrt{4288} \approx 65.5
\]
Now substituting back into the formula for \( x \):
\[
x = \frac{64 \pm 65.5}{2}
\]
This gives us two potential solutions:
1. \( x = \frac{64 + 65.5}{2} = \frac{129.5}{2} \approx 64.75 \)
2. \( x = \frac{64 - 65.5}{2} = \frac{-1.5}{2} \approx -0.75 \) (not physically meaningful since distance cannot be negative)
So the distance from the child where the ball strikes the ground is
\[
\approx 64.75 \text{ feet}.
\]
### Summary of Answers:
- (a) 3 feet
- (b) 67 feet
- (c) 64.75 feet
### (a) Height when the ball leaves the child's hand
When the ball leaves the child's hand, \( x = 0 \).
Substituting \( x = 0 \) into the height equation:
\[
y(0) = -\frac{1}{16}(0)^2 + 4(0) + 3 = 3
\]
**So, the height when the ball leaves the child's hand is 3 feet.**
### (b) Maximum height of the ball
The maximum height of a parabolic equation in the form \( y = ax^2 + bx + c \) occurs at the vertex, which can be found using the formula:
\[
x = -\frac{b}{2a}
\]
Here, \( a = -\frac{1}{16} \) and \( b = 4 \).
Calculating the x-coordinate of the vertex:
\[
x = -\frac{4}{2 \cdot -\frac{1}{16}} = -\frac{4}{-\frac{1}{8}} = 4 \cdot 8 = 32
\]
Now substituting \( x = 32 \) back into the height equation to find the maximum height:
\[
y(32) = -\frac{1}{16}(32)^2 + 4(32) + 3
\]
Calculating each term:
\[
y(32) = -\frac{1}{16}(1024) + 128 + 3 = -64 + 128 + 3 = 67
\]
**Thus, the maximum height of the ball is 67 feet.**
### (c) How far from the child does the ball strike the ground?
To find out where the ball strikes the ground, we set \( y = 0 \) and solve for \( x \):
\[
0 = -\frac{1}{16}x^2 + 4x + 3
\]
Multiplying through by -16 to eliminate fractions:
\[
0 = x^2 - 64x - 48
\]
Now we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -64 \), and \( c = -48 \).
Calculating the discriminant:
\[
b^2 - 4ac = (-64)^2 - 4(1)(-48) = 4096 + 192 = 4288
\]
Now applying the quadratic formula:
\[
x = \frac{64 \pm \sqrt{4288}}{2}
\]
Calculating \( \sqrt{4288} \):
\[
\sqrt{4288} \approx 65.5
\]
Now substituting back into the formula for \( x \):
\[
x = \frac{64 \pm 65.5}{2}
\]
This gives us two potential solutions:
1. \( x = \frac{64 + 65.5}{2} = \frac{129.5}{2} \approx 64.75 \)
2. \( x = \frac{64 - 65.5}{2} = \frac{-1.5}{2} \approx -0.75 \) (not physically meaningful since distance cannot be negative)
So the distance from the child where the ball strikes the ground is
\[
\approx 64.75 \text{ feet}.
\]
### Summary of Answers:
- (a) 3 feet
- (b) 67 feet
- (c) 64.75 feet