Question

A ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity of 80 feet per second. The distance s​ (in feet) of the ball from the ground after t seconds is s = 96 + 80 t - 16 t².

Answer 1

(a) After 6 seconds ball strike he ground.

(b) After 5 seconds ball pass the top of the building on its way down.

Explanation:

Consider the complete question is "A ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity of 80 feet per second. The distance S (in feet )of the ball from the ground, after t seconds is s=96+80t-16t^2.

(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?"

The given equation is

`s=96+80t-16^2`

(a)

The distances​ of the ball from the ground is 0 when it strike the ground.

Substitute s=0 in the above equation.

`0=96+80t-16^2`

Divide both sides by 16.

`0=6+5t-t^2`

`0=6+6t-t-t^2`

`0=6(1+t)-t(1+t)`

`0=(1+t)(6-t)`

Using zero product property we get

`t=-1,6`

The value of time can no be negative. So, t=6.

Therefore, after 6 seconds ball strike he ground.

(b)

We need to find the time after which the ball pass the top of the building on its way down.

Substitute s=96 in the above equation.

`96=96+80t-16^2`

`0=16t(5-t)`

Using zero product property we get

`t=0,5`

Therefore, after 5 seconds ball pass the top of the building on its way down.