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)

The path of the projectile resembles an upside-down parabola, where it initially ascends as it progresses downrange, reaches a peak at its maximum height, and then descends back toward the ground. This is typical of projectile motion, showing that the object climbs, reaches a point of maximum elevation, and then falls as gravitational forces take over. To find the maximum height of the projectile, you can use the vertex formula for a parabola given by \( h = -0.2x^2 + 135x + 10 \). The x-coordinate of the vertex, which gives the maximum height, can be found using \( x = -\frac{b}{2a} \) where \( a = -0.2 \) and \( b = 135 \). Plugging in the values, \( x = -\frac{135}{2 \cdot -0.2} = 337.5 \). Now substituting this back into the height formula gives \( h = -0.2(337.5)^2 + 135(337.5) + 10 \), allowing you to calculate the maximum height achieved by the projectile.