A ball is thrown upward and outward from a height of 5 feet. The height of the ball, \( f(x) \), in feet, can be modeled by \( f(x)=-0.1 x^{2}+0.6 x+5 \) where \( x \) is the ball's horizontal distance, in feet, from where it was thrown. Use this model to solve parts (a) through (c) a. What is the maximum height of the ball and how far from where it was thrown does this occur? The maximum height is \( \square \) feet, which occurs \( \square \) feet from the point of release. (Round to the nearest tenth as needed.)

To find the maximum height of the ball, you can use the vertex formula for a quadratic function. The x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \), where \( a = -0.1 \) and \( b = 0.6 \). Plugging in these values gives you: \[ x = -\frac{0.6}{2 \times -0.1} = 3 \] Now, substituting \( x = 3 \) back into the height equation \( f(x) \): \[ f(3) = -0.1(3^2) + 0.6(3) + 5 = -0.1(9) + 1.8 + 5 \] \[ f(3) = -0.9 + 1.8 + 5 = 5.9 \] Thus, the maximum height of the ball is \( 5.9 \) feet, which occurs \( 3 \) feet from the point of release. The math behind projectile motion can be traced all the way back to ancient civilizations, where early thinkers like Aristotle hypothesized about the motions of objects. Fast forward to the Renaissance, and figures like Galileo introduced principles of physics that laid the foundation for our modern understanding—such as the idea that projectiles follow a curved path due to the influence of gravity. In a real-world scenario, you might find this kind of motion in sports, such as basketball shots or when throwing a ball during a game. Understanding the trajectory helps players calculate the optimal angle and force needed for making successful shots. Engineers also utilize these principles when designing roller coasters to ensure thrilling yet safe experiences!