ournal 10 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 \( f \) 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 ur down 4 shape, starting at a neight of IOft ris to a peak and then clescending back towards ground cis it travels horizentailey 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 range of the projectile is the distance it travels horizontally. How far down range do 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 (\(-0.2\)), the parabola opens downwards. - This means that the projectile will rise to a maximum height and then fall back down as it travels horizontally. - The projectile starts at a height of \( 10 \) feet, rises to a peak, and then descends back towards the ground. ### 2. Finding the Maximum Height To find the maximum height of the projectile, we can use the vertex formula for a parabola given by \( x = -\frac{b}{2a} \), where \( a \) and \( b \) are the coefficients from the quadratic equation \( h = ax^2 + bx + c \). In our case: - \( a = -0.2 \) - \( b = 135 \) Let's calculate the \( x \) value at which the maximum height occurs: \[ x = -\frac{135}{2 \times -0.2} \] Now, we will calculate this value. Calculate the value by following steps: - step0: Calculate: \(\frac{-135}{\left(2\left(-0.2\right)\right)}\) - step1: Remove the parentheses: \(\frac{-135}{2\left(-0.2\right)}\) - step2: Multiply the numbers: \(\frac{-135}{-0.4}\) - step3: Convert the expressions: \(\frac{-135}{-\frac{2}{5}}\) - step4: Multiply by the reciprocal: \(-135\left(-\frac{5}{2}\right)\) - step5: Use the rules for multiplication and division: \(135\times \frac{5}{2}\) - step6: Multiply: \(\frac{135\times 5}{2}\) - step7: Multiply: \(\frac{675}{2}\) The maximum height occurs at \( x = 337.5 \) feet. Now, let's find the maximum height by substituting \( x = 337.5 \) back into the height equation: \[ h = -0.2(337.5)^2 + 135(337.5) + 10 \] Let's calculate this value. 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: \(\frac{91165}{4}\) The maximum height of the projectile is \( 22,791.25 \) feet, which confirms the value provided in the problem. ### 3. Finding the Distance to Reach a Height of 110 ft To find how far downrange the projectile must travel to reach a height of \( 110 \) feet, we set the height equation equal to \( 110 \): \[ -0.2x^2 + 135x + 10 = 110 \] Rearranging this equation gives: \[ -0.2x^2 + 135x - 100 = 0 \] Now, we can solve this quadratic equation for \( x \). Let's use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where: - \( a = -0.2 \) - \( b = 135 \) - \( c = -100 \) Let's calculate the discriminant \( b^2 - 4ac \) first. Calculate the value by following steps: - step0: Calculate: \(135^{2}-4\left(-0.2\right)\left(-100\right)\) - step1: Multiply: \(135^{2}-80\) - step2: Evaluate the power: \(18225-80\) - step3: Subtract the numbers: \(18145\) The discriminant \( b^2 - 4ac \) is \( 18145 \). Now, we can use the quadratic formula to find the values of \( x \): \[ x = \frac{-135 \pm \sqrt{18145}}{2 \times -0.2} \] Let's calculate the two possible values for \( x \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(-135+\sqrt{18145}\right)}{\left(2\left(-0.2\right)\right)}\) - step1: Remove the parentheses: \(\frac{-135+\sqrt{18145}}{2\left(-0.2\right)}\) - step2: Multiply the numbers: \(\frac{-135+\sqrt{18145}}{-0.4}\) - step3: Convert the expressions: \(\frac{-135+\sqrt{18145}}{-\frac{2}{5}}\) - step4: Multiply by the reciprocal: \(\left(-135+\sqrt{18145}\right)\left(-\frac{5}{2}\right)\) - step5: Use the rules for multiplication and division: \(\left(135-\sqrt{18145}\right)\times \frac{5}{2}\) - step6: Multiply: \(\frac{\left(135-\sqrt{18145}\right)\times 5}{2}\) - step7: Multiply: \(\frac{675-5\sqrt{18145}}{2}\) Calculate or simplify the expression \( (-135 - \sqrt(18145))/(2*-0.2) \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(-135-\sqrt{18145}\right)}{\left(2\left(-0.2\right)\right)}\) - step1: Remove the parentheses: \(\frac{-135-\sqrt{18145}}{2\left(-0.2\right)}\) - step2: Multiply the numbers: \(\frac{-135-\sqrt{18145}}{-0.4}\) - step3: Convert the expressions: \(\frac{-135-\sqrt{18145}}{-\frac{2}{5}}\) - step4: Multiply by the reciprocal: \(\left(-135-\sqrt{18145}\right)\left(-\frac{5}{2}\right)\) - step5: Use the rules for multiplication and division: \(\left(135+\sqrt{18145}\right)\times \frac{5}{2}\) - step6: Multiply: \(\frac{\left(135+\sqrt{18145}\right)\times 5}{2}\) - step7: Multiply: \(\frac{675+5\sqrt{18145}}{2}\) The two possible values for \( x \) are approximately: 1. \( x \approx 0.74 \) (which is not relevant in this context since it represents a negative height) 2. \( x \approx 674.26 \) Thus, the projectile will have to travel approximately \( 674.26 \) feet downrange to reach a height of \( 110 \) feet. ### Summary of Results - The shape of the projectile's path resembles a downward-opening parabola. - The maximum height of the projectile is \( 22,791.25 \) feet. - The distance downrange to reach a height of \( 110 \) feet is approximately \( 674.26 \) feet.