Answer:
c.) .5 seconds
Step-by-step explanation:
y = -16t^2 + 4t + 2
If we want to know when the object hits the groupd, that means y, the height, must be 0.
0 = -16t^2 + 4t + 2
Now we can solve. We can take a factor of 2 out of every term to make it easier on the brain :,)
0 = 2(-8t^2 + 2t + 1)
0/2 = -8t^2 + 2t + 1
0 = -8t^2 + 2t + 1
Now we can use the quadratic formula to get the solutions. (Remember that it's [tex]x = \frac{-b +- \sqrt{b^{2}-4ac} }{2a}[/tex], with the standard form of a quadratic being ax^2 + bx + c.) You could also factor, if you like that sort of thing.
x = 1/2, -1/4
Now we know we can't have negative seconds, so the answer is 1/2 seconds, or c.) .5 seconds.
