A ball is thrown vertically upward. After $t$ seconds, its height $h$ (in feet) is given by the function $h(t)=100 t-16 t^{2}$. What is the maximum height that the ball will reach?
Do not round your answer.

Step 1 ：The maximum height of the ball is the maximum value of the function \(h(t)\). This occurs at the vertex of the parabola represented by the function.

Step 2 ：The \(t\)-coordinate of the vertex of a parabola given by \(f(t) = at^2 + bt + c\) is \(-\frac{b}{2a}\).

Step 3 ：In this case, \(a = -16\) and \(b = 100\), so the time at which the maximum height is reached is \(-\frac{100}{2(-16)}\).

Step 4 ：We can then substitute this value back into the function to find the maximum height.

Step 5 ：\(t = 3.125\)

Step 6 ：\(h = -16*t^2 + 100*t\)

Step 7 ：\(h_{max} = 156.250000000000\)

Step 8 ：Final Answer: The maximum height that the ball will reach is \(\boxed{156.25}\) feet.