Answer:
The diver will hit the water at 1.5 seconds
Step-by-step explanation:
Given
[tex]h= -16(t-1.5)(t+1)[/tex]
Required (Missing from the question)
When will the diver hit the water?
To do this, we simply solve for t
When the diver hits the water, the height is 0 (at that point)
So, substitute 0 for h in [tex]h= -16(t-1.5)(t+1)[/tex]
[tex]0= -16(t-1.5)(t+1)[/tex]
Divide both sides by -16
[tex]\frac{0}{-16} =\frac{-16(t-1.5)(t+1)}{-16}[/tex]
[tex]\frac{0}{-16} =(t-1.5)(t+1)[/tex]
[tex]0 =(t-1.5)(t+1)[/tex]
[tex](t-1.5)(t+1)=0[/tex]
Split
[tex]t - 1.5 = 0[/tex] or [tex]t + 1 = 0[/tex]
Solve for t
[tex]t = 1.5[/tex] or [tex]t = -1[/tex]
But time (t) can not be negative.
So:
[tex]t = 1.5[/tex]
Hence, the diver will hit the water at 1.5 seconds
