A projectile is launched from a height of \( 10^{\prime} \) and follows a path described by the formula \( h=-.2 x^{2}+135 x+10 \) where \( x \) is the distance the projectile has travelled downrange and \( h \) is the height of the projectile. 1. Describe in your own words what the shape of path this would look like? \( (A) \) The path of the projectile will resemble an upside down 4 shape, starting at a neight of 10 ft rising to a peak and then descending back towaras the groundl as it tracues horizentaity 2. Find the max height of this projectile.(B) The projectile reccnes a maximum heiaht 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 approximately 674.26 ft clownrange to reach 110 ft . 4. The range of the projectile is the distance it travels horizontally. How far down range does the projectile travel. (D)

Let's break down the problem step by step. ### 1. Shape of the Path The path of the projectile is described by the quadratic equation \( h = -0.2x^2 + 135x + 10 \). Since the coefficient of \( x^2 \) is negative, this indicates that the parabola opens downwards. In simpler terms, the projectile will rise to a maximum height and then fall back down, resembling an upside-down "U" shape. It starts at a height of 10 feet, rises to a peak, and then descends back toward the ground as it travels horizontally. ### 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. The x-coordinate of the vertex for a quadratic equation in the form \( ax^2 + bx + c \) can be found using the formula: \[ x = -\frac{b}{2a} \] Here, \( a = -0.2 \) and \( b = 135 \). Let's calculate the x-coordinate of the vertex: \[ x = -\frac{135}{2 \times -0.2} = -\frac{135}{-0.4} = 337.5 \] Now, we can substitute \( x = 337.5 \) back into the height equation to find the maximum height \( h \): \[ h = -0.2(337.5)^2 + 135(337.5) + 10 \] Now, let's calculate this value. ### 3. Finding the Height at 110 ft To find how far downrange the projectile travels to reach a height of 110 ft, we set \( h = 110 \) and solve for \( x \): \[ 110 = -0.2x^2 + 135x + 10 \] Rearranging gives us: \[ -0.2x^2 + 135x - 100 = 0 \] Now, we can solve this quadratic equation for \( x \). ### 4. Finding the Range of the Projectile The range of the projectile is the distance it travels horizontally when it reaches the ground level (height \( h = 0 \)). We set \( h = 0 \) and solve for \( x \): \[ 0 = -0.2x^2 + 135x + 10 \] Now, let's perform the calculations for the maximum height, the distance to reach 110 ft, and the range of the projectile. I'll calculate these values now. Calculate the value by following steps: - step0: Calculate: \(-0.2\times 337.5^{2}+135\times 337.5+10\) - step1: Convert the expressions: \(-0.2\left(\frac{675}{2}\right)^{2}+135\times 337.5+10\) - step2: Multiply the numbers: \(-\frac{91125}{2^{2}}+135\times 337.5+10\) - step3: Multiply the numbers: \(-\frac{91125}{2^{2}}+45562.5+10\) - step4: Add the numbers: \(-\frac{91125}{2^{2}}+45572.5\) - step5: Convert the expressions: \(-\frac{91125}{2^{2}}+\frac{91145}{2}\) - step6: Reduce fractions to a common denominator: \(-\frac{91125}{2^{2}}+\frac{91145\times 2}{2\times 2}\) - step7: Multiply the numbers: \(-\frac{91125}{2^{2}}+\frac{91145\times 2}{4}\) - step8: Rewrite the expression: \(-\frac{91125}{2^{2}}+\frac{91145\times 2}{2^{2}}\) - step9: Transform the expression: \(\frac{-91125+91145\times 2}{2^{2}}\) - step10: Multiply the numbers: \(\frac{-91125+182290}{2^{2}}\) - step11: Add the numbers: \(\frac{91165}{2^{2}}\) - step12: Evaluate the power: \(\frac{91165}{4}\) Solve the equation \( -0.2x^2 + 135x + 10 = 0 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(-0.2x^{2}+135x+10=0\) - step1: Convert the decimal into a fraction: \(-\frac{1}{5}x^{2}+135x+10=0\) - step2: Multiply both sides: \(\frac{1}{5}x^{2}-135x-10=0\) - step3: Multiply both sides: \(5\left(\frac{1}{5}x^{2}-135x-10\right)=5\times 0\) - step4: Calculate: \(x^{2}-675x-50=0\) - step5: Solve using the quadratic formula: \(x=\frac{675\pm \sqrt{\left(-675\right)^{2}-4\left(-50\right)}}{2}\) - step6: Simplify the expression: \(x=\frac{675\pm \sqrt{675^{2}+200}}{2}\) - step7: Simplify the expression: \(x=\frac{675\pm 5\sqrt{18233}}{2}\) - step8: Separate into possible cases: \(\begin{align}&x=\frac{675+5\sqrt{18233}}{2}\\&x=\frac{675-5\sqrt{18233}}{2}\end{align}\) - step9: Rewrite: \(x_{1}=\frac{675-5\sqrt{18233}}{2},x_{2}=\frac{675+5\sqrt{18233}}{2}\) Solve the equation \( -0.2x^2 + 135x - 100 = 0 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(-0.2x^{2}+135x-100=0\) - step1: Convert the decimal into a fraction: \(-\frac{1}{5}x^{2}+135x-100=0\) - step2: Multiply both sides: \(\frac{1}{5}x^{2}-135x+100=0\) - step3: Multiply both sides: \(5\left(\frac{1}{5}x^{2}-135x+100\right)=5\times 0\) - step4: Calculate: \(x^{2}-675x+500=0\) - step5: Solve using the quadratic formula: \(x=\frac{675\pm \sqrt{\left(-675\right)^{2}-4\times 500}}{2}\) - step6: Simplify the expression: \(x=\frac{675\pm \sqrt{675^{2}-2000}}{2}\) - step7: Simplify the expression: \(x=\frac{675\pm 5\sqrt{18145}}{2}\) - step8: Separate into possible cases: \(\begin{align}&x=\frac{675+5\sqrt{18145}}{2}\\&x=\frac{675-5\sqrt{18145}}{2}\end{align}\) - step9: Rewrite: \(x_{1}=\frac{675-5\sqrt{18145}}{2},x_{2}=\frac{675+5\sqrt{18145}}{2}\) ### 2. Maximum Height The maximum height of the projectile is calculated to be: \[ h = 22791.25 \text{ feet} \] ### 3. Distance to Reach 110 ft To find the distance \( x \) when the height \( h = 110 \) ft, we solved the equation: \[ -0.2x^2 + 135x - 100 = 0 \] The solutions are: \[ x_{1} = \frac{675 - 5\sqrt{18145}}{2}, \quad x_{2} = \frac{675 + 5\sqrt{18145}}{2} \] ### 4. Range of the Projectile To find the range when the projectile reaches the ground level (height \( h = 0 \)), we solved the equation: \[ -0.2x^2 + 135x + 10 = 0 \] The solutions are: \[ x_{1} = \frac{675 - 5\sqrt{18233}}{2}, \quad x_{2} = \frac{675 + 5\sqrt{18233}}{2} \] Now, let's calculate the numerical values for the distances \( x_1 \) and \( x_2 \) for both the height of 110 ft and the range. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(675+5\sqrt{18145}\right)}{2}\) - step1: Remove the parentheses: \(\frac{675+5\sqrt{18145}}{2}\) Calculate or simplify the expression \( (675 + 5*\sqrt(18233))/2 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(675+5\sqrt{18233}\right)}{2}\) - step1: Remove the parentheses: \(\frac{675+5\sqrt{18233}}{2}\) Calculate or simplify the expression \( (675 - 5*\sqrt(18233))/2 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(675-5\sqrt{18233}\right)}{2}\) - step1: Remove the parentheses: \(\frac{675-5\sqrt{18233}}{2}\) Calculate or simplify the expression \( (675 - 5*\sqrt(18145))/2 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(675-5\sqrt{18145}\right)}{2}\) - step1: Remove the parentheses: \(\frac{675-5\sqrt{18145}}{2}\) ### Summary of Results 1. **Maximum Height**: The projectile reaches a maximum height of approximately \( 22791.25 \) feet. 2. **Distance to Reach 110 ft**: - The projectile travels approximately \( 674.26 \) feet downrange to reach a height of \( 110 \) ft (using the positive root \( x_2 \approx 674.26 \)). - The negative root \( x_1 \approx 0.74 \) feet is not physically meaningful in this context. 3. **Range of the Projectile**: - The projectile travels approximately \( 675.07 \) feet downrange when it reaches the ground level (using the positive root \( x_2 \approx 675.07 \)). - The negative root \( x_1 \approx -0.07 \) feet is also not physically meaningful. ### Final Answers - **Maximum Height**: \( 22791.25 \) feet - **Distance to 110 ft**: \( 674.26 \) feet - **Range of the Projectile**: \( 675.07 \) feet