Answer:
0.64s
Step-by-step explanation:
Hello!
To solve this problem we must use the equation to determine for what time (t), the ball touches the ground that is when h = 0
[tex]h =-16t^2 + 4t + 4\\for h=0\\0=-16t^2 + 4t + 4\\[/tex]
Now we have a quadratic equation, which has two answers, for this we use the following equation
a = number that accompanies the variable t ^ 2
=-16
b = number that accompanies the variable t
=4
c = independent number of t=4
[tex]t=\frac{-b+\sqrt{b^2-4ac} }{2a}[/tex]
solving
[tex]t=\frac{-4(-+)\sqrt[2]{4^2-4(-16)(4)} }{2(-16)}[/tex]
remember that when the equation is quadratic it has two answers, one with the sign (+), and another with the sign (-), you choose the positive result since it is impossible for the time to be negative
t1=0.64s
t2=-0.39s
answer=0.64s
