A ball is thrown upward from a height of 256 feet above the ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time $t$ is $v(t)=96-32 t$ feet per second. a) Find $s(t)$, the function giving the height of the ball at time $t$. b) How long will the ball take to reach the ground? c) How high will the ball go? a) $s(t)=\square$ b) The ball will reach the ground after $\square$ seconds. c) The ball will go $\square$ feet high.

Step 1 ：The height of the ball at time \(t\) is given by the integral of the velocity function \(v(t)\). Therefore, we have: \[s(t) = \int v(t) dt = \int (96 - 32t) dt = 96t - 16t^2 + C\].

Step 2 ：The constant of integration \(C\) can be determined by the initial condition that the ball is thrown from a height of 256 feet. So when \(t=0\), \(s(0) = 256\). Therefore, \(C = 256\).

Step 3 ：So, \[s(t) = 96t - 16t^2 + 256\].

Step 4 ：The ball will reach the ground when \(s(t) = 0\). So we need to solve the equation \[96t - 16t^2 + 256 = 0\].

Step 5 ：We solve it using the quadratic formula: \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-96 \pm \sqrt{96^2 - 4*(-16)*256}}{2*(-16)}\].

Step 6 ：We discard the negative root because time cannot be negative, so \[t = \frac{-96 + \sqrt{96^2 + 16384}}{-32} = 6\] seconds.

Step 7 ：The ball will reach its maximum height when the velocity is zero, i.e., when \(v(t) = 0\). So we need to solve the equation \[96 - 32t = 0\], which gives \(t = 3\) seconds.

Step 8 ：Substituting this into the height function gives the maximum height: \[s(3) = 96*3 - 16*3^2 + 256 = 288\] feet.

Step 9 ：\(\boxed{s(t) = 96t - 16t^2 + 256}\)

Step 10 ：\(\boxed{\text{The ball will reach the ground after 6 seconds.}}\)

Step 11 ：\(\boxed{\text{The ball will go 288 feet high.}}\)