Solve the problem.
-A ball is thrown vertically upward from the top of a building 144 feet tall with an initial velocity of 128 feet per second. The distance s (in feet)of the ball from the ground after t seconds is s = 144 + 128t - 16t2. After how many seconds will the ball pass the top of the building on its way down?
A)10 sec
B)8 sec
C)7 sec
D)144 sec

Question

Quizplus · Accepted Answer

The ball passes the top of the building on its way down when its height is equal to the height of the building, which is 144 feet. Setting s equal to 144 gives the equation 144 = 144 + 128t - 16t^2. Simplifying, we get 0 = 128t - 16t^2. Factoring out 16t gives 16t(8 - t) = 0, which gives t = 0 (when the ball is initially at the top) and t = 8 (when the ball is coming back down). Since we're interested in when the ball is coming down, the answer is 8 seconds.

Solve the Problem. -A Ball Is Thrown Vertically Upward from the Top of the Top