The interval notation for the time(s) the ball is lower than the bridge will be: Â [tex](2, 3.384...][/tex]
Explanation
Height of the bridge is 75 ft.
The height of the object can be modeled by the equation: Â [tex]s(t) = -16t^2 + 32t + 75[/tex]
The object is lower than the bridge means, the height of the object will be less than 75 ft and when it reaches the ground then it's height will be 0.
When [tex]s(t)<0[/tex] , then.......
[tex]-16t^2+32t+75<75\\ \\ -16t^2+32t<0\\ \\ -16t(t-2)<0[/tex]
As [tex]t[/tex] can't be less than 0, so [tex]t-2 > 0[/tex] or [tex]t>2[/tex]
When [tex]s(t)=0[/tex] , then........
[tex]-16t^2 + 32t + 75=0 \\ \\ t=\frac{-32\pm \sqrt{32^2-4(-16)(75)}}{2(-16)}\\ \\ t=\frac{-32 \pm \sqrt{1024+4800}}{-32}\\ \\ t=\frac{-32\pm \sqrt{5824}}{-32}\\ \\ t= -1.384..., t=3.384...[/tex]
(Negative value is ignored)
So, the interval notation for the time(s) the ball is lower than the bridge will be: Â [tex](2, 3.384...][/tex]
[tex](-\infty ,0)[/tex] is not included in the solution as it means [tex]t<0[/tex] , but time can't be in negative. The time, [tex]t[/tex] was 0 at the time of trowing the object. So, [tex]t[/tex] can't be less than 0.
Answer:
Time(s) the ball is lower than the bridge is (2, 3.385)
Here (-∞,0) is not included because time cannot be negative, it is always positive.
Step-by-step explanation:
We have equation rock given by -16t²+32t+75.
At t = 0 the height is 75 m.
We need to find after how much time the height is less than 75.
That is
        -16t²+32t+75 ≤ 75
        -16t²+32t  ≤ 0
         16t² - 32 t ≥ 0
          t² - 2t ≥ 0
          t (t-2) ≥ 0
   That is   t ≥ 2
We also need to check when the stone reaches ground, that is
-16t²+32t+75 = 0
On solving we will get
t = 3.385 s and t = -1.385 seconds ( not possible)
So at time 3.385 seconds stone reaches ground.
So, Time(s) the ball is lower than the bridge is (2, 3.385)
Here (-∞,0) is not included because time cannot be negative, it is always positive.
