Journal 10 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 down 4 shape, starting at a neight of loft ris to a peak and then clescending back towards ground cis it travels horizentalty. 2. Find the max height of this projectile.(B)

**Step 1. Understanding the Shape of the Path** The height \( h(x) \) is given by the quadratic equation \[ h(x) = -0.2x^2 + 135x + 10. \] Because the coefficient of \( x^2 \) (which is \(-0.2\)) is negative, the graph is a parabola that opens downward. This means the path of the projectile starts at a height of \( 10 \) feet, rises to a maximum height at the vertex of the parabola, and then falls back down as \( x \) increases. **Step 2. Finding the \( x \)-coordinate of the Vertex** For a quadratic equation of the form \[ h(x) = ax^2 + bx + c, \] the \( x \)-coordinate of the vertex (which gives the point of maximum height when \( a < 0 \)) is found by \[ x = -\frac{b}{2a}. \] Here, \( a = -0.2 \) and \( b = 135 \). Thus, \[ x = -\frac{135}{2(-0.2)} = \frac{135}{0.4}. \] Calculating, \[ x = 337.5. \] **Step 3. Calculating the Maximum Height** Substitute \( x = 337.5 \) back into the equation for \( h(x) \): \[ h(337.5) = -0.2(337.5)^2 + 135(337.5) + 10. \] First, compute \( (337.5)^2 \): \[ (337.5)^2 = 113906.25. \] Multiply by \(-0.2\): \[ -0.2 \times 113906.25 = -22781.25. \] Next, calculate \( 135 \times 337.5 \): \[ 135 \times 337.5 = 45562.5. \] Now, sum the computed values with the constant: \[ h(337.5) = -22781.25 + 45562.5 + 10 = 22791.25. \] Thus, the maximum height reached by the projectile is \[ 22791.25 \text{ feet}. \]