Answer:
D) 14 seconds
Step-by-step explanation:
First we will plug 500 in for y:
500 = -4.9t² + 120t
We want to set this equal to 0 in order to solve it; to do this, subtract 500 from each side:
500-500 = -4.9t² + 120t - 500
0 = -4.9t²+120t-500
Our values for a, b and c are:
a = -4.9; b = 120; c = -500
We will use the quadratic formula to solve this. Â This will give us the two times that the object is at exactly 500 meters. Â The difference between these two times will tell us when the object is at or above 500 meters.
The quadratic formula is:
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Using our values for a, b and c,
[tex]x=\frac{-120\pm \sqrt{120^2-4(-4.9)(-500)}}{2(-4.9)}\\\\=\frac{-120\pm \sqrt{14400-9800}}{-9.8}\\\\=\frac{-120\pm \sqrt{4600}}{-9.8}\\\\=\frac{-120\pm 67.8233}{-9.8}\\\\=\frac{-120+67.8233}{-9.8}\text{ or }\frac{-120-67.8233}{-9.8}\\\\=\frac{-57.1767}{-9.8}\text{ or }\frac{-187.8233}{-9.8}\\\\=5.3242\text{ or }19.1656[/tex]
The two times the object is at exactly 500 meters above the ground are at 5 seconds and 19 seconds. Â This means the amount of time it is at or above 500 meters is
19-5 = 14 seconds.
