A ball is thrown vertically upward from the top of a building 160 feet tall with an initial velocity of 48 feet per second. The distance s (in feet) of the ball from the ground after t seconds is $$ s(t)=160+48 t-16 t^{2} $$, Tuestion 4 of 16 sest: 16 point(s) possible (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 $$ \square $$ seconds. (s) (b)

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**(a) Finding when the ball strikes the ground**

The ball strikes the ground when its height $$ s(t) = 0 $$. The height function is given by:
$$
s(t) = 160 + 48t - 16t^2.
$$
Setting $$ s(t) = 0 $$:
$$
160 + 48t - 16t^2 = 0.
$$
Divide the entire equation by 16 to simplify:
$$
10 + 3t - t^2 = 0.
$$
Rearrange the equation into standard quadratic form:
$$
-t^2 + 3t + 10 = 0.
$$
Multiply the equation by $$-1$$ to make the coefficient of $$t^2$$ positive:
$$
t^2 - 3t - 10 = 0.
$$
Now, apply the quadratic formula:
$$
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
$$
where $$a = 1$$, $$b = -3$$, and $$c = -10$$. Substitute these values:
$$
t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)} = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm \sqrt{49}}{2}.
$$
Since $$\sqrt{49} = 7$$, the solutions are:
$$
t = \frac{3 + 7}{2} = 5 \quad \text{or} \quad t = \frac{3 - 7}{2} = -2.
$$
Since time cannot be negative, the ball strikes the ground after:
$$
\boxed{5 \text{ seconds}}.
$$

**(b) Finding when the ball passes the top of the building on its way down**

The top of the building is at 160 feet. We need to find when:
$$
s(t) = 160.
$$
Substitute $$s(t)$$:
$$
160 + 48t - 16t^2 = 160.
$$
Subtract 160 from both sides:
$$
48t - 16t^2 = 0.
$$
Factor out $$16t$$:
$$
16t(3 - t) = 0.
$$
Set each factor to zero:
$$
16t = 0 \quad \Longrightarrow \quad t = 0,
$$
$$
3 - t = 0 \quad \Longrightarrow \quad t = 3.
$$
The value $$t = 0$$ corresponds to the initial time, when the ball is thrown. Thus, on its way down, the ball passes the top of the building at:
$$
\boxed{3 \text{ seconds}}.
$$
**(a) The ball strikes the ground after 5 seconds.** **(b) The ball passes the top of the building on its way down after 3 seconds.**

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