Answer:
t1= 1.1475 sec
t2=9.48 sec
Step-by-step explanation:
This question asking for the time when an object reach a specific height. I will explain what the formula stand for to make this easier.
h= height = 174feet
v0= initial velocity = 170 feet/s
t= time
h0= initial height= ground level = 0
The question have predetermined formula for the height, so we should use it instead of formula that involve gravity value.
h= -16[tex]t^{2}[/tex] +v0t + h0t
174 feet = -16 [tex]t^{2}[/tex] + 170 feet/s x t + 0 x t
-16 [tex]t^{2}[/tex] + 170 feet/s x t - 174 feet = 0
8 [tex]t^{2}[/tex] - 85 feet/s x t + 87 feet = 0
This equation will be hard to solve since its value in fraction, so lets use the quadratic formula
[tex]x= \frac{-b±\sqrt{ b^2-4ac}}{2a}[/tex]
[tex]t= \frac{(85)±\sqrt{ -85^2-4(8)(87)}}{2(8)}[/tex]
[tex]t= \frac{85±\ 66.64}{16}[/tex]
t1= [tex]\frac{85-\ 66.64}{16}[/tex]
t1= 1.1475
t2= [tex]\frac{85+\ 66.64}{16}[/tex]
t2=9.48
