Question

A ball is thrown upward from the top of a building 200 feet tall. The height of the ball is described by the function h(t) = -16r^2 + 12+ + 200, where t is the time in seconds and t = 0 corresponds to the moment the ball is thrown. (a) Determine for which value of the ball reaches the maximum height and determine this maximum height.

Answer 1

The maximum height of the ball thrown upward from a 200-foot building is reached at 0.375 seconds, and the maximum height attained is 203 feet.

Step-by-step explanation:

The equation provided, h(t) = -16t^2 + 12t + 200, is a quadratic equation that describes the height (h) of a ball at any time (t) after it has been thrown upward from the top of a 200-foot building. To determine the time when the ball reaches its maximum height, we need to find the vertex of the parabola described by the quadratic equation. The vertex form of a parabola given by y = ax^2 + bx + c is h = k when t = -b/(2a). In this case, a = -16 and b = 12, so the time at which the ball reaches its maximum height is t = -12/(2(-16)) = 0.375 seconds. The maximum height can be found by substituting this time back into the height equation, which yields h(0.375) = -16(0.375)^2 + 12(0.375) + 200 = 203 feet.