A person standing close to the edge on top of a 64-foot building throws a ball vertically upward. The quadratic function $h(t)=-16 t^{2}+120 t+64$ models the ball's height about the ground, $h(t)$, in feet, $t$ seconds after it was thrown. a) What is the maximum height of the ball? feet b) How many seconds does it take until the ball hits the ground? seconds Question Help: $\square$ Video Submit Question

Step 1 ：The maximum height of the ball can be found by finding the vertex of the parabola represented by the quadratic function. The x-coordinate of the vertex of a parabola given by the equation \(y = ax^2 + bx + c\) is \(-\frac{b}{2a}\). In this case, \(a = -16\) and \(b = 120\), so the time at which the ball reaches its maximum height is \(-\frac{120}{2(-16)}\).

Step 2 ：Substituting this value into the equation will give the maximum height.

Step 3 ：The maximum height of the ball is \(\boxed{289}\) feet.