Answer:
t = 2.5 s
Step-by-step explanation:
Suppose you have a water-balloon launcher. The balloon is 3 ft high when it leaves the launcher. Its equation is :
[tex]16t^2 + 38.8t + 3 =0[/tex]
The above equation is a quadratic equation. The general equation is :
[tex]ax^2+bx+c=0[/tex]
Here, a = 16, b = 38.8 and c = 3
The solution of quadratic equation is given by :
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}\\\\t=\dfrac{-38.8\pm \sqrt{(38.8)^2-4\times (-16)\times 3} }{2\times (-16)}\\\\t=\dfrac{-38.8+\sqrt{(38.8)^{2}-4\times-16\times3}}{-2\times16}, \dfrac{-38.8-\sqrt{(38.8)^{2}-4\times-16\times3}}{-2\times16}\\\\t=-0.075\ s,2.5\ s[/tex]
So, at t = 2.5 s the balloon is in air.
