Aball 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=80+64t-16t^2.a) after how many seconds does the ball strike the ground? b) after how many seconds does the ball pass the top of the building on its way down?

a)

b)

Explanation:

The knowable variables are the initial hight and initial velocity

The equation that describes the motion of the ball is:

If we want to know the time that takes the ball to hit the ground, we need to calculate it by doing s=0 that is the final hight.

a) Solving for t, we are going to have two answers

a=-16

b=64

c=80

or

Since time can not be negative the answer is t=6.378s

b) To find the time that takes the ball to pass the top of the building on its way down, we must find how much does it move too

First of all, we need to find the maximum hight and how much time does it take to reach it:

at maximum point the velocity is 0

Solving for t

Now, we must know how much distance does it take to reach maximum point

So, the ball pass the top of the building on its way down at 160 ft

Solving for t

or

Since the time that the ball reaches maximum point is almost t=2s that answer can not be possible, so the answer is t=3.333s for the ball to go up and down, passing the top of the building

(a) 5 s

(b) 4s

Explanation:

height of building, s = 80 feet

The equation of motion is given by

(a) Let it takes t time to reach the ground. A it reach the ground, s = 0

So, put s = 0 in the given equation, we get

(t + 1)(t - 5) = 0

t = -1 s, t = 5 s

As time cannot be negative, so t = 5 s

Thus, the time taken by the ball to reach the ground is 5 s.

(b) Let it crosses the building in t second, s put s = 80 feet in the above equation, we get

t(4 - t)= 0

t = 0s or t  4 s

So, it takes 4 second to cross the building.