For this case we have that the equation that describes the height is:
[tex] h = -16t ^ 2 + vt + c
[/tex]
Where,
v: initial speed
c: initial height
Substituting values we have:
[tex] h = -16t ^ 2 + 138t + 55
[/tex]
Then, to find the time, we use the quadratic formula:
[tex] t =\frac{-b +/-\sqrt{b^2 - 4ac} }{2a} [/tex]
Substituting values we have:
[tex] t =\frac{-138 +/-\sqrt{138^2 - 4(-16)(55)} }{2(-16)} [/tex]
Rewriting:
[tex] t =\frac{-138 +/-\sqrt{19044 + 3520} }{-32} [/tex]
[tex] t =\frac{-138 +/-\sqrt{22564}}{-32} [/tex]
[tex] t =\frac{-138 +/-150.21}{-32} [/tex]
We discard the negative root because we want to find the time.
We have then:
[tex] t =\frac{-138 -150.21}{-32} [/tex]
[tex] t =\frac{-288.21}{-32}
t=9 [/tex]
Answer:
D. [tex] 0 = -16t^2 + 138t + 55; 9 s
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