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)

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.