Question

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+128−16t^2
After how many seconds does the ball strike the ground?
a) 2
b) 4
c) 6
d) 8

After how many seconds will the ball pass the top of the building on its way down?
a) 2
b) 4
c) 6
d) 8

Answer 1

The ball strikes the ground after 6 seconds and passes the top of the building on its way down after 2 seconds.

Step-by-step explanation:

To find the time when the ball strikes the ground, we need to set the distance equation s = 144 + 128t - 16t^2 equal to zero since the ball will be on the ground at that time. We can solve this quadratic equation using the quadratic formula, which gives us two solutions: t = 6 seconds and t = 8 seconds. Since time t = 0 is when the ball is released at the top of the building, we are only interested in the positive root, which is t = 6 seconds. Therefore, the ball strikes the ground after 6 seconds.

To find the time when the ball passes the top of the building on its way down, we need to set the distance equation s = 144 + 128t - 16t^2 equal to 144, the height of the building. Again, we can use the quadratic formula to solve this equation. The positive root is the one we are interested in since time t = 0 is when the ball is released at the top of the building. Therefore, the ball passes the top of the building on its way down after 2 seconds.