Lorraine hits a golf ball into the air. It lands 60 feet away from her. At its highest point, the ball is 30 feet in the air.

Question

Desmos Link

(6 points)
This scenario would best be represented by a
equation.

The ball lands 60 feet away from her would represent the
. Half of this distance would be axis of symmetry. That would make the vertex

The ball starts at how many feet away from Lorraine?
So another point on the graph would be

Using the information, the only value that is not given and needs to be found is the

Question 10
What is the equation for this scenario?

What is the height of the golf ball when it is 15 feet away from Lorrain?

(2 points)
The equation is f(x) =

(answer using fractions and do not include grouping symbols)

The height of the ball when it is 15 feet away from Lorraine is
feet in the air. (round to the nearest tenth)

Question 11
10.

While hiking, Julia kicked a small rock off a ledge that was meters above a crevasse. Graph the equation f(x) = −4.9x2+9
. Use the graph to determine when the rock will hit the bottom of the crevasse, 2 meters below ground level. Round your answer to the nearest tenth of a second.

Desmos Link

(1 point)
The rock will hit the bottom of the crevasse in
seconds.

Answer 1

Let's break down the questions one at a time.

Question 10

1. Determine the equation that models the scenario

Given that the golf ball lands 60 feet away from Lorraine and reaches its maximum height of 30 feet, we can model the situation as a quadratic equation in vertex form:

\[ f(x) = a(x - h)^2 + k \]

Where:

• \((h, k)\) is the vertex of the parabola.
• \(h\) is the horizontal distance from Lorraine to the vertex (which is the axis of symmetry).
• \(k\) is the maximum height.

Finding \(h\): The ball lands at 60 feet, so it travels 30 feet to reach the vertex at its highest point (30 feet high). Thus:

\[ h = \frac{60}{2} = 30 \] \[ k = 30 \]

So the vertex is at \( (30, 30) \).

Finding \(a\): We know the ball lands back to the ground at \(y = 0\) when \(x = 60\). So we can use this point \( (60, 0) \) to find \(a\).

Substituting into the vertex form equation: \[ 0 = a(60 - 30)^2 + 30 \] \[ 0 = a(30)^2 + 30 \] \[ 0 = 900a + 30 \] \[ 900a = -30 \] \[ a = -\frac{1}{30} \]

So the equation becomes: \[ f(x) = -\frac{1}{30}(x - 30)^2 + 30 \]

This can be expanded to: \[ f(x) = -\frac{1}{30}(x^2 - 60x + 900) + 30 \] \[ f(x) = -\frac{1}{30}x^2 + 2x + 30 \]

So the answer would be: \[ f(x) = -\frac{1}{30}x^2 + 2x \]

2. Finding height at \(15\) feet away from Lorraine

To find the height of the ball when it is \(15\) feet away from Lorraine, we would substitute \(x = 15\) into the equation:

\[ f(15) = -\frac{1}{30}(15)^2 + 2(15) + 30 \] \[ = -\frac{1}{30}(225) + 30 + 30 \] \[ = -7.5 + 30 + 30 \] \[ = 52.5 \]

So, the height of the ball when it is \(15\) feet away from Lorraine is 52.5 feet.

Question 11

For the equation given for Julia's rock:

\[ f(x) = -4.9x^2 + 9 \]

We want to find when the rock will hit the bottom of the crevasse, which is \(2\) meters below ground level, or \(y = -2\). We can set the equation equal to \(-2\):

\[ -4.9x^2 + 9 = -2 \]

Rearranging gives us:

\[ -4.9x^2 + 9 + 2 = 0 \] \[ -4.9x^2 + 11 = 0 \] \[ 4.9x^2 = 11 \] \[ x^2 = \frac{11}{4.9} \] \[ x = \sqrt{\frac{11}{4.9}} \approx \sqrt{2.2449} \approx 1.5 , \text{(to nearest tenth)} \]

Thus, the rock will hit the bottom of the crevasse in approximately 1.5 seconds.