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