The height 

y

 (in feet) of a ball thrown by a child is

y=-1 1/2x^2 +6x+5

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? 
  
 feet


The equation of the height \( y \) of the ball, given as:

\[
y = -\frac{3}{2}x^2 + 6x + 5
\]

where \( x \) is the horizontal distance from the point of release, allows us to analyze the height of the ball when it leaves the child’s hand, find the maximum height, and determine where it strikes the ground. Let's go through each part.

---

### (a) Height of the Ball When It Leaves the Child's Hand
The ball leaves the child's hand when \( x = 0 \) (initial position). Substitute \( x = 0 \) into the equation:

\[
y = -\frac{3}{2}(0)^2 + 6(0) + 5 = 5
\]

**Answer: The ball is 5 feet high when it leaves the child’s hand.**

---

### (b) Maximum Height of the Ball
Since this is a quadratic equation opening downward (the coefficient of \( x^2 \) is negative), the maximum height occurs at the vertex of the parabola. For a quadratic equation of the form \( y = ax^2 + bx + c \), the \( x \)-coordinate of the vertex is given by:

\[
x = -\frac{b}{2a}
\]

Here, \( a = -\frac{3}{2} \) and \( b = 6 \):

\[
x = -\frac{6}{2 \cdot -\frac{3}{2}} = -\frac{6}{-3} = 2
\]

Substitute \( x = 2 \) back into the equation to find \( y \):

\[
y = -\frac{3}{2}(2)^2 + 6(2) + 5
\]
\[
y = -\frac{3}{2} \cdot 4 + 12 + 5
\]
\[
y = -6 + 12 + 5 = 11
\]

**Answer: The maximum height of the ball is 11 feet.**

---

### (c) Distance From the Child When the Ball Strikes the Ground
The ball strikes the ground when \( y = 0 \). So, we need to solve:

\[
0 = -\frac{3}{2}x^2 + 6x + 5
\]

To clear the fraction, multiply every term by 2:

\[
0 = -3x^2 + 12x + 10
\]

Now, use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = -3 \), \( b = 12 \), and \( c = 10 \):

\[
x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot (-3) \cdot 10}}{2 \cdot -3}
\]
\[
x = \frac{-12 \pm \sqrt{144 + 120}}{-6}
\]
\[
x = \frac{-12 \pm \sqrt{264}}{-6}
\]
\[
x = \frac{-12 \pm 16.25}{-6}
\]

This yields two solutions:

1. \( x = \frac{-12 + 16.25}{-6} \approx -0.71 \) (not relevant as distance cannot be negative)
2. \( x = \frac{-12 - 16.25}{-6} \approx 4.37 \)

**Answer: The ball strikes the ground approximately 4.37 feet from the child.**

---

Would you like more details on any part of this solution?

1. How can we verify the height when the ball leaves the child's hand?
2. What does the vertex of a parabola represent in physics-related motion problems?
3. How can we apply the quadratic formula to solve real-world problems?
4. Why is only one solution relevant when finding the distance the ball strikes the ground?
5. What changes if the initial height of the throw changes?

**Tip:** The vertex of a parabola in a motion equation gives the point of maximum (or minimum) value, which is helpful in problems involving height, speed, or distance.

The height y (in feet) of a ball thrown by a child is y = -1.5x^2 + 6x + 5 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?
(b) What is the maximum height of the ball?
(c) How far from the child does the ball strike the ground?

Find the height, maximum height, and distance of a ball thrown by a child using the quadratic function y = -1.5x^2 + 6x + 5. This solution covers the quadratic formula and vertex calculation, ideal for Grades 9-11 students.

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