A ball is thrown vertically upward from the top of a building 128 feet tall with an initial velocity of 32 feet per second. The distance \( s \) (in feet) of the ball from the ground after \( t \) seconds is \( s(t)=128+32 t-16 t^{2} \). (a) After how many seconds does the ball strike the ground? (b) After how many seconds will the ball pass the top of the building on its way down? (a) After \( \square \) seconds the ball strikes the ground. (b) The ball passes the top of the building on its way down after seconds.

To find when the ball strikes the ground, we need to set the distance function \( s(t) \) equal to 0 and solve for \( t \). Given: \[ s(t) = 128 + 32t - 16t^2 \] Setting \( s(t) = 0 \): \[ 0 = 128 + 32t - 16t^2 \] Now, we can solve this quadratic equation to find the time when the ball strikes the ground. Solve the quadratic equation by following steps: - step0: Solve by factoring: \(0=128+32t-16t^{2}\) - step1: Swap the sides: \(128+32t-16t^{2}=0\) - step2: Factor the expression: \(16\left(4-t\right)\left(2+t\right)=0\) - step3: Divide the terms: \(\left(4-t\right)\left(2+t\right)=0\) - step4: Separate into possible cases: \(\begin{align}&4-t=0\\&2+t=0\end{align}\) - step5: Solve the equation: \(\begin{align}&t=4\\&t=-2\end{align}\) - step6: Rewrite: \(t_{1}=-2,t_{2}=4\) The solutions to the quadratic equation are \( t_1 = -2 \) and \( t_2 = 4 \). Since time cannot be negative, the ball strikes the ground after 4 seconds. Now, to find when the ball passes the top of the building on its way down, we need to set the distance function \( s(t) \) equal to 128 (the height of the building) and solve for \( t \). Setting \( s(t) = 128 \): \[ 128 = 128 + 32t - 16t^2 \] Simplifying the equation: \[ 0 = 32t - 16t^2 \] Factoring out \( t \): \[ 0 = t(32 - 16t) \] Setting each factor equal to 0: \[ t = 0 \text{ or } 32 - 16t = 0 \] Solving for \( t \): \[ t = 0 \text{ or } t = 2 \] Since the ball passes the top of the building on its way down after 2 seconds, the answer is 2 seconds.