A ball is thrown vertically upward. After $t$ seconds, its height $h$ (in feet) is given by the function $h(t)=72 t-16 t^{2}$. After how long will it reach its maximum height? Do not round your answer.

Step 1 ：Given the height function \(h(t)=72 t-16 t^{2}\), we need to find the time at which the ball reaches its maximum height.

Step 2 ：The maximum height is reached when the velocity of the ball is zero. The velocity of the ball is given by the derivative of the height function.

Step 3 ：Find the derivative of the height function: \(h'(t) = 72 - 32t\).

Step 4 ：Set the derivative equal to zero and solve for \(t\): \(72 - 32t = 0\).

Step 5 ：Solving the equation gives \(t = \frac{9}{4}\).

Step 6 ：Final Answer: The ball will reach its maximum height after \(\boxed{\frac{9}{4}}\) seconds.