Answer:
t =0.06s and t=9.94 s
Step-by-step explanation:
The equation that models the height of the rocket is
[tex]h(t) = - 16 {t}^{2} + 160t[/tex]
when the rocket is at height of 9 fret above the ground, then
[tex]h(t) = 9[/tex]
This implies that:
[tex] - 16 {t}^{2} + 160t = 9[/tex]
Rewrite in standard quadratic form.
[tex]- 16 {t}^{2} + 160t - 9 = 0[/tex]
The solution is given by
[tex]t = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} [/tex]
[tex]t = \frac{ - 160 \pm \sqrt{ {160}^{2} - 4 \times - 16 \times - 9} }{2 \times - 9} [/tex]
[tex]t = 5 + \frac{\sqrt{391} }{4} \: or \: t = 5 - \frac{\sqrt{391} }{4} [/tex]
[tex]t = 9.94 \: or \: .06[/tex]
The rocket reach the height of 9 feet at times t=0.06s and t=9.94 seconds
