A ball is thrown upward with an initial velocity of $48 \mathrm{ft} / \mathrm{sec}$ from a height $864 \mathrm{ft}$. Its height s, in feet, after $\mathrm{t}$ seconds is given by $s=-16 t^{2}+48 t+864$. After how long will the ball reach the ground? The ball will reach the ground in seconds

Step 1 ：The ball will reach the ground when its height is 0. So, we need to solve the equation \(-16 t^{2}+48 t+864 = 0\) for \(t\).

Step 2 ：The solutions to the equation are -6 and 9. However, time cannot be negative, so we discard -6.

Step 3 ：Therefore, the ball will reach the ground after 9 seconds.

Step 4 ：Final Answer: The ball will reach the ground after \(\boxed{9}\) seconds.