Answer :
Answer:
[tex] 16t^2 -38.8 t -3 =0[/tex]
And we can use the quadratic formula given by:
[tex] t = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]
Where:
[tex] a = 16, b=-38.8, c = -3[/tex]
And replacing we got:
[tex] t = \frac{38.8 \pm \sqrt{(-38.8)^2 -4(16)(-3)}}{2*16} [/tex]
And after solve we got two solutions:
[tex] t_1 = 2.5 s[/tex]
And [tex] t_2 =-0.075 s[/tex]
Since the time can't be negative the correct option for this case would be [tex] t =2.5 s[/tex]
Step-by-step explanation:
For this case we have the following function for the height:
[tex] h(t) = -16t^2 +38.8t +3[/tex]
And we want to find how many seconds t that the balloon is in the air since is released from 3ft above, so we want to find the time t in order to h(t) =0
[tex] 0= -16t^2 +38.8t +3[/tex]
We can rewrite the last expression like this:
[tex] 16t^2 -38.8 t -3 =0[/tex]
And we can use the quadratic formula given by:
[tex] t = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]
Where:
[tex] a = 16, b=-38.8, c = -3[/tex]
And replacing we got:
[tex] t = \frac{38.8 \pm \sqrt{(-38.8)^2 -4(16)(-3)}}{2*16} [/tex]
And after solve we got two solutions:
[tex] t_1 = 2.5 s[/tex]
And [tex] t_2 =-0.075 s[/tex]
Since the time can't be negative the correct option for this case would be [tex] t =2.5 s[/tex]