Question

a ball is thrown vertically upward from the top of a building 64 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 = 64 + 48t - 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 how many seconds does the ball strike the ground? t = seconds (b) after how many seconds will the ball pass the top of the building on its way down? t = seconds

Explanation:

Step1: Set up equation for part (a)

The ball strikes the ground when $s = 0$. So we set $64 + 48t-16t^{2}=0$. Divide through by - 16 to simplify: $t^{2}-3t - 4=0$.

Step2: Factor the quadratic equation

We factor $t^{2}-3t - 4$ as $(t - 4)(t+1)=0$.

Step3: Solve for t

Using the zero - product property, if $(t - 4)(t + 1)=0$, then $t-4=0$ or $t + 1=0$. So $t=4$ or $t=-1$. Since time cannot be negative, the ball strikes the ground at $t = 4$ seconds.

Step4: Set up equation for part (b)

The ball passes the top of the building on its way down when $s = 64$. So we set $64+48t-16t^{2}=64$. Subtract 64 from both sides to get $48t-16t^{2}=0$. Factor out $16t$: $16t(3 - t)=0$.

Step5: Solve for t

Using the zero - product property, if $16t(3 - t)=0$, then $16t=0$ or $3 - t=0$. The solution $t = 0$ corresponds to the time of throwing. The other solution is $t = 3$ seconds when the ball passes the top of the building on its way down.

Answer:

(a) $t = 4$ seconds
(b) $t = 3$ seconds