Answer:
Step-by-step explanation:
The quadratic is
[tex]s(t)=-4.9t^2+12t+1.8[/tex]
where s(t) is the height of the ball after a certain amount of time goes by. If we are looking for how long til it hits the ground, we know 2 things from that question. First we know that the height of an object on the ground is 0, and we also know that t is the time that the ball is in the air. Because the height of the ball when it's on the ground is 0, we will set s(t) equal to 0 and factor the quadratic to solve for t, the time the ball is in the air. Throw it into the quadratic formula to find those times.
[tex]t=\frac{-12+/-\sqrt{144-4(-4.9)(1.8)} }{2(-4.9)}[/tex] which will simplify to
[tex]t=\frac{-12+/-\sqrt{179.28} }{-9.8}[/tex]
The 2 solutions we get from that are that
[tex]t=\frac{-12+\sqrt{179.28} }{-9.8}=-.1417906 sec[/tex] and
[tex]t=\frac{-12-\sqrt{179.28} }{-9.8}=2.59077sec[/tex]
Now, the 2 things in math that will never be negative are distances and time, so we will disregard the negative time and go with t = 2.59 seconds. Not sure to where you need to round.
