a calculator is allowed for this question. a ...
a calculator is allowed for this question. a golf ball is hit so that it travels a horizontal distance of 440 feet and reaches a maximum height of 190 feet. a. determine a quadratic equation that models the path of the golf ball, assuming it starts at the origin. (4 points) round the coefficient of x² to the nearest ten - thousandths place, and round the coefficient of x to the nearest thousandths place. b. using your understanding of parametric equations for projectile motion, what is the angle the ball takes off? show your steps. (4 points) a calculator may be used for this problem. to write the equation, select the “insert” drop - down option in the menu below and use “√x equation “ to type your answer when you need to incorporate symbols, equations, or other math expressions. edit view insert format tools table 12pt paragraph b i u a t²
Answer
# Answer:
## A.
The quadratic - equation of a parabola with vertex - form \(y=a(x - h)^2 + k\), where the vertex \((h,k)\) is the maximum point of the parabola. The ball starts at the origin \((0,0)\) and the maximum point is \((h,k)=(220,190)\) (since the maximum occurs at the mid - point of the horizontal range, and the range is \(440\)).
Substitute \((x,y)=(0,0)\) and \((h,k)=(220,190)\) into \(y=a(x - h)^2 + k\):
\(0=a(0 - 220)^2+190\)
\(0 = 48400a+190\)
\(48400a=-190\)
\(a=-\frac{190}{48400}\approx - 0.0039\)
The equation in vertex - form is \(y=-0.0039(x - 220)^2+190\)
Expand it: \(y=-0.0039(x^{2}-440x + 48400)+190\)
\(y=-0.0039x^{2}+1.716x-188.76 + 190\)
\(y=-0.0039x^{2}+1.716x + 1.24\)
## B.
The parametric equations for projectile motion are \(x = v_0\cos\theta t\) and \(y=v_0\sin\theta t-\frac{1}{2}gt^{2}\), where \(g = 32\mathrm{ft/s}^2\).
At the maximum height, the vertical velocity \(v_y = v_0\sin\theta−gt = 0\), so \(t=\frac{v_0\sin\theta}{g}\)
The maximum height \(y_{max}=v_0\sin\theta\cdot\frac{v_0\sin\theta}{g}-\frac{1}{2}g(\frac{v_0\sin\theta}{g})^{2}=\frac{v_0^{2}\sin^{2}\theta}{2g}\)
The range \(R=\frac{2v_0^{2}\sin\theta\cos\theta}{g}\)
We know that \(y_{max}=190\) and \(R = 440\)
From \(y_{max}=\frac{v_0^{2}\sin^{2}\theta}{2g}\), we have \(v_0^{2}=\frac{2gy_{max}}{\sin^{2}\theta}\)
Substitute \(v_0^{2}\) into the range formula \(R=\frac{2v_0^{2}\sin\theta\cos\theta}{g}\):
\(R=\frac{2\cdot\frac{2gy_{max}}{\sin^{2}\theta}\cdot\sin\theta\cos\theta}{g}\)
\(R = 4y_{max}\cot\theta\)
\(\cot\theta=\frac{R}{4y_{max}}\)
Substitute \(R = 440\) and \(y_{max}=190\)
\(\cot\theta=\frac{440}{4\times190}=\frac{440}{760}\approx0.579\)
\(\theta=\arctan(\frac{1}{\cot\theta})\)
\(\theta=\arctan(\frac{760}{440})\approx\arctan(1.727)\approx59.9^{\circ}\approx60^{\circ}\)
# Explanation:
## A.
### Step1: Identify vertex - form
Use \(y=a(x - h)^2 + k\) with \((h,k)=(220,190)\)
### Step2: Find \(a\)
Substitute \((0,0)\) into \(y=a(x - 220)^2+190\)
### Step3: Expand
Expand \(y=-0.0039(x - 220)^2+190\)
## B.
### Step1: Recall parametric equations
Use \(x = v_0\cos\theta t\) and \(y=v_0\sin\theta t-\frac{1}{2}gt^{2}\)
### Step2: Find \(t\) at max height
Set \(v_y = 0\) to get \(t=\frac{v_0\sin\theta}{g}\)
### Step3: Express \(v_0^{2}\) from max - height
\(v_0^{2}=\frac{2gy_{max}}{\sin^{2}\theta}\)
### Step4: Substitute into range formula
Substitute \(v_0^{2}\) into \(R=\frac{2v_0^{2}\sin\theta\cos\theta}{g}\)
### Step5: Solve for \(\theta\)
Find \(\theta\) from \(\cot\theta=\frac{R}{4y_{max}}\)