A ball is thrown vertically upward with an initial velocity of 64 feet per second. The distance $s$ (in feet) of the ball from the ground after $t$ seconds is $s=64 t-16 t^{2}$. (a) At what time $t$ will the ball strike the ground? (b) For what time $t$ is the ball more than 28 feet above the ground?

Step 1 ：We are given that a ball is thrown vertically upward with an initial velocity of 64 feet per second. The distance $s$ (in feet) of the ball from the ground after $t$ seconds is given by the equation $s=64 t-16 t^{2}$.

Step 2 ：For part (a), we need to find the time $t$ when the ball strikes the ground. This happens when the distance $s$ is zero. So we solve the equation $64t - 16t^2 = 0$ for $t$.

Step 3 ：Solving the equation gives us two solutions: $t=0$ and $t=4$ seconds. However, since $t=0$ is the initial time when the ball is thrown, the ball will strike the ground at $t=4$ seconds.

Step 4 ：So, the ball will strike the ground at $t=\boxed{4}$ seconds.

Step 5 ：For part (b), we need to find the time $t$ when the ball is more than 28 feet above the ground. This happens when $64t - 16t^2 > 28$. So we solve this inequality for $t$.

Step 6 ：Solving the inequality gives us the interval $t \in \left(\frac{1}{2}, \frac{7}{2}\right)$ seconds.

Step 7 ：So, the ball is more than 28 feet above the ground for $t$ in the interval $\boxed{\left(\frac{1}{2}, \frac{7}{2}\right)}$ seconds.