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