A ball is thrown vertically upward. After $t$ seconds, its height $h$ (in feet) is given by the function $h(t)=48 t-16 t^{2}$. After how long will it reach its maximum height? Do not round your answer. ? Time: $\square$ seconds $\times 5$

Step 1 ：The maximum height of the ball is reached when the velocity of the ball is 0. The velocity is the derivative of the height function. So, we need to find the derivative of the height function and set it equal to 0 to find the time at which the ball reaches its maximum height.

Step 2 ：Given the height function \(h(t) = -16t^2 + 48t\), we find its derivative \(h'(t) = 48 - 32t\).

Step 3 ：Setting \(h'(t) = 0\), we solve for \(t\) to get \(t = \frac{3}{2}\).

Step 4 ：Final Answer: The ball will reach its maximum height after \(\boxed{1.5}\) seconds.