Question

a ball is thrown vertically upward from the top of a building 80 feet tall with an initial velocity of 64 feet per second. the distance s (in feet) of the ball from the ground after t seconds is s(t)=80 + 64t-16t². (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 seconds the ball strikes the ground. (b) the ball passes the top of the building on its way down after seconds.

Explanation:

Step1: Identify the height - time formula

The height formula for an object in vertical - motion under the influence of gravity is $s(t)=s_0 + v_0t-\frac{1}{2}gt^2$, where $s_0$ is the initial height, $v_0$ is the initial velocity, and $g$ is the acceleration due to gravity. In the English system of units, $g = 32$ ft/s². Here, $s_0 = 80$ ft and $v_0=64$ ft/s, so $s(t)=80 + 64t-16t^2$.

Step2: Find the time when the ball strikes the ground

When the ball strikes the ground, $s(t)=0$. So we set up the quadratic equation $-16t^2 + 64t + 80=0$. Divide through by - 16 to simplify: $t^2-4t - 5 = 0$.

Step3: Solve the quadratic equation

Factor the quadratic equation $t^2-4t - 5=(t - 5)(t+1)=0$. Using the zero - product property, $t - 5 = 0$ or $t + 1=0$. So $t = 5$ or $t=-1$. Since time cannot be negative in this context, the ball strikes the ground at $t = 5$ seconds.

Step4: Find the time when the ball reaches the top of the building on the way down

The initial height of the ball is $s_0 = 80$ ft. We set $s(t)=80$ in the equation $s(t)=80 + 64t-16t^2$. So $80=80 + 64t-16t^2$. Subtract 80 from both sides to get $0=64t-16t^2$. Factor out $16t$: $16t(4 - t)=0$. This gives two solutions: $t = 0$ (corresponds to the time of launch) and $t = 4$ (corresponds to the time when the ball comes back to the same height on the way down).

Answer:

(a) 5
(b) 4