Question

A ball that is thrown upward follows the path h(t)=-t^2+8t+32, where h(t) represents the height of the ball above the ground at time t, and t represents the time since the ball was thrown. How high above the ground did the ball reach? show work

Answer 1

Explanation:

To find the maximum height of the ball, we need to find the vertex of the parabolic function h(t) = -t^2 + 8t + 32. The vertex represents the highest point of the parabolic curve.

The x-coordinate of the vertex is given by the formula x = -b/2a, where a and b are the coefficients of the quadratic equation. In this case, a = -1 and b = 8, so we have:

x = -b/2a = -8/(2(-1)) = 4

Therefore, the vertex occurs at time t = 4.

To find the y-coordinate of the vertex, we evaluate the function at t = 4:

h(4) = -4^2 + 8(4) + 32 = 16 + 32 = 48

Therefore, the ball reaches a maximum height of 48 feet above the ground.