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

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Answer

Final Answer: The ball will reach the ground after $\boxed{9}$ seconds.

Steps

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.