April shoots an arrow upward into the air at a speed of 64 feet per second from a platform that is 27 feet high. The height of the arrow is given by the function h(t)=-16 t^{2}+64 t+27, where t is the time is seconds. What is the maximum height of the arrow?

Step 1 ：Given that April shoots an arrow upward into the air at a speed of 64 feet per second from a platform that is 27 feet high. The height of the arrow is given by the function \(h(t)=-16 t^{2}+64 t+27\), where t is the time is seconds.

Step 2 ：The maximum height of the arrow can be found by finding the vertex of the parabola represented by the function \(h(t)=-16 t^{2}+64 t+27\).

Step 3 ：The x-coordinate of the vertex of a parabola given by the equation \(y = ax^2 + bx + c\) is given by \(-b/2a\). In this case, \(a = -16\), \(b = 64\).

Step 4 ：So, the time at which the arrow reaches its maximum height is \(t = -b/2a\).

Step 5 ：We can then substitute this value of \(t\) into the function \(h(t)\) to find the maximum height.

Step 6 ：Substituting the values, we get \(a = -16\), \(b = 64\), \(c = 27\), \(t = 2.0\), and \(max\_height = 91.0\).

Step 7 ：Final Answer: The maximum height of the arrow is \(\boxed{91}\) feet.