A projectile is launched from a height of $$ 10^{\prime} $$ and follows a path described by the formula $$ h=-.2 x^{2}+135 x $$ of the projectile. 1. Describe in your own words what the shape of path this would look like? $$ (A) $$ The Dath of the projectile will resemble an down 4 shape, starting at a neight of IOft ris to a peak and then cescending back toward grouncl as it travels harizentally. 2. Find the max height of this projectile.(B) The projectile recenes a maximum heic Of 22,791,25 feet. 3. How far down range will the projectile have to travel to reach a height of 110 ft ?(C) The projectile travels cipproximately clowrange to reach 110 ft . The range of the projectile is the distance it travels horizontally. How far down range does projectile travel. (D)

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UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### 1. Shape of the Path
The path described by the formula $$ h = -0.2x^2 + 135x $$ is a quadratic equation. The negative coefficient of $$ x^2 $$ indicates that the parabola opens downwards.

In simpler terms, the projectile will rise to a peak height and then fall back down as it moves horizontally. It starts at a height of $$ 10 $$ feet, rises to a maximum height, and then descends back toward the ground. This is characteristic of projectile motion.

### 2. Finding the Maximum Height
To find the maximum height of the projectile, we need to determine the vertex of the parabola represented by the equation $$ h = -0.2x^2 + 135x $$. The x-coordinate of the vertex can be found using the formula:

$$
x = -\frac{b}{2a}
$$

where $$ a = -0.2 $$ and $$ b = 135 $$.

Let's calculate the x-coordinate of the vertex.

### 3. Finding the Distance to Reach a Height of 110 ft
To find how far downrange the projectile travels to reach a height of $$ 110 $$ ft, we need to set $$ h = 110 $$ in the equation and solve for $$ x $$:

$$
-0.2x^2 + 135x = 110
$$

This can be rearranged to form a standard quadratic equation:

$$
-0.2x^2 + 135x - 110 = 0
$$

Now, let's perform the calculations for both the maximum height and the distance to reach $$ 110 $$ ft.

### Performing Calculations
I will calculate the x-coordinate of the vertex and solve the quadratic equation for $$ h = 110 $$ ft.

Let's proceed with these calculations.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x=-\frac{135}{2\left(-0.2\right)}$$
- step1: Simplify:
  $$x=\frac{675}{2}$$
Solve the equation $$ -0.2x^2 + 135x - 110 = 0 $$.
Solve the quadratic equation by following steps:
- step0: Solve using the quadratic formula:
  $$-0.2x^{2}+135x-110=0$$
- step1: Convert the decimal into a fraction:
  $$-\frac{1}{5}x^{2}+135x-110=0$$
- step2: Multiply both sides:
  $$\frac{1}{5}x^{2}-135x+110=0$$
- step3: Multiply both sides:
  $$5\left(\frac{1}{5}x^{2}-135x+110\right)=5\times 0$$
- step4: Calculate:
  $$x^{2}-675x+550=0$$
- step5: Solve using the quadratic formula:
  $$x=\frac{675\pm \sqrt{\left(-675\right)^{2}-4\times 550}}{2}$$
- step6: Simplify the expression:
  $$x=\frac{675\pm \sqrt{675^{2}-2200}}{2}$$
- step7: Simplify the expression:
  $$x=\frac{675\pm 5\sqrt{18137}}{2}$$
- step8: Separate into possible cases:
  $$\begin{align}&x=\frac{675+5\sqrt{18137}}{2}\&x=\frac{675-5\sqrt{18137}}{2}\end{align}$$
- step9: Rewrite:
  $$x_{1}=\frac{675-5\sqrt{18137}}{2},x_{2}=\frac{675+5\sqrt{18137}}{2}$$
### 2. Maximum Height Calculation
The x-coordinate of the vertex, which gives us the maximum height, is calculated as follows:

$$
x = \frac{675}{2} = 337.5
$$

Now, we can substitute $$ x = 337.5 $$ back into the height equation to find the maximum height:

$$
h = -0.2(337.5)^2 + 135(337.5)
$$

Let's calculate this maximum height.
Solve the equation by following steps:
- step0: Solve for $$h$$:
  $$h=-0.2\times 337.5^{2}+135\times 337.5$$
- step1: Simplify:
  $$h=\frac{91125}{2^{2}}$$
- step2: Evaluate the power:
  $$h=\frac{91125}{4}$$
The maximum height of the projectile is:

$$
h = 22781.25 \text{ feet}
$$

### 3. Finding the Distance to Reach a Height of 110 ft
We previously set up the quadratic equation to find the distance $$ x $$ when the height $$ h $$ is $$ 110 $$ ft:

$$
-0.2x^2 + 135x - 110 = 0
$$

The solutions to this equation are:

$$
x_{1} = \frac{675 - 5\sqrt{18137}}{2}, \quad x_{2} = \frac{675 + 5\sqrt{18137}}{2}
$$

Now, let's calculate the approximate values of $$ x_1 $$ and $$ x_2 $$ to find how far downrange the projectile travels to reach a height of $$ 110 $$ ft.
Calculate the value by following steps:
- step0: Calculate:
  $$675-5\sqrt{18137}$$
Calculate or simplify the expression $$ 675 + 5*\sqrt(18137) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$675+5\sqrt{18137}$$
The approximate values for the solutions to the quadratic equation are:

1. $$ x_1 \approx 1.63 $$
2. $$ x_2 \approx 1348.37 $$

### Summary of Results
- **Maximum Height (B)**: The maximum height of the projectile is $$ 22781.25 $$ feet.
- **Distance to Reach 110 ft (C)**: The projectile reaches a height of $$ 110 $$ ft at approximately $$ 1.63 $$ feet and $$ 1348.37 $$ feet downrange.

### 4. Total Downrange Distance (D)
The total downrange distance the projectile travels is the distance at which it reaches the ground after reaching its maximum height. The two distances calculated for $$ 110 $$ ft indicate that the projectile will be at that height twice during its flight: once on the way up and once on the way down.

Thus, the projectile travels approximately $$ 1.63 $$ feet and $$ 1348.37 $$ feet downrange to reach a height of $$ 110 $$ ft.
The projectile reaches a maximum height of 22,781.25 feet. It travels approximately 1.63 feet and 1,348.37 feet downrange to reach a height of 110 feet.

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