Answer :
Given:
The distance s (in feet) of the ball from the ground after t seconds is
[tex]s=64t-16t^2[/tex]
To find:
(a) At what time t will the ball strike the ground?
(b) For what time t is the ball more than 48 feet above the ground?
Solution:
(a)
We have,
[tex]s=64t-16t^2[/tex]
Substitute s=0, to find the time t when the ball strike the ground.
[tex]0=64t-16t^2[/tex]
[tex]0=16t(4-t)[/tex]
Using zero product property, we get
[tex]16t=0\Rightarrow t=0[/tex]
[tex]4-t=0\Rightarrow t=4[/tex]
Therefore, the ball strike the ground in initial condition (t = 0) and after 4 seconds (t = 4).
(b)
Now, s > 48, to find the time t when the ball will be more than 48 feet above the ground.
[tex]64t-16t^2\geq 48[/tex]
[tex]0> 16t^2-64t+48[/tex]
Divide both sides by 16.
[tex]0> t^2-4t+3[/tex]
[tex]0> t^2-t-3t+3[/tex]
[tex]0> t(t-1)-3(t-1)[/tex]
[tex]0>(t-1)(t-3)[/tex]
Related equation is [tex](t-1)(t-3)=0[/tex]. Zeroes are t=1,3. These two number divide the number line is three parts. (-∞,1),(1,3),(3,∞)
[tex]0>(t-1)(t-3)[/tex] inequality is true for only (1,3).
It is only possible when t lies in the interval (1,3).
Therefore, the ball will be more than 48 feet above the ground between 1 and 3 seconds.