Question

How the heck do I solve this? Please help T^T

Answer 1

Answer:

Step-by-step explanation:

The quadratic is

$s(t)=-4.9t^2+12t+1.8$

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.

$t=\frac{-12+/-\sqrt{144-4(-4.9)(1.8)} }{2(-4.9)}$ which will simplify to

$t=\frac{-12+/-\sqrt{179.28} }{-9.8}$

The 2 solutions we get from that are that

$t=\frac{-12+\sqrt{179.28} }{-9.8}=-.1417906 sec$ and

$t=\frac{-12-\sqrt{179.28} }{-9.8}=2.59077sec$

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.