A ball is thrown upward with an initial velocity of $32 \mathrm{ft} / \mathrm{sec}$ from a height $768 \mathrm{ft}$. Its height $\mathrm{s}$, in feet, after $t$ seconds is given by $s=-16 t^{2}+32 t+768$. After how long will the ball reach the ground?
Answer
Final Answer: The ball will reach the ground after \(\boxed{8}\) seconds.
Step 1 ：A ball is thrown upward with an initial velocity of 32 ft/sec from a height of 768 ft. Its height s, in feet, after t seconds is given by \(s=-16 t^{2}+32 t+768\). We need to find out after how long will the ball reach the ground.
Step 2 ：The ball reaches the ground when its height is 0. So, we need to solve the equation \(-16t^2 + 32t + 768 = 0\) for t.
Step 3 ：The solutions to the equation are -6 and 8. However, time cannot be negative, so we discard -6.
Step 4 ：Final Answer: The ball will reach the ground after \(\boxed{8}\) seconds.
