Suppose you have a water balloon launcher. The balloon is 3ft high when it leaves the launcher. Use the equation 0=-16t^2+38.8t+3 to find the number of seconds t that the balloon is in the air.

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]