Answer fully : ) 30 pts
an object is launched from ground level with an initial velocity of 120 meters per second. for how long is the object at or above 500 meters (rounded to the nearest second)?

the equation that models the path of the object is y = -4.9t^2 + 120t.
a) 11 seconds
b) 12 seconds
c) 13 seconds
d) 14 seconds

D)14 seconds

Step-by-step explanation:

y = -4.9t^2 + 120t

We want to find out when y = 500

500 = -4.9t^2 + 120t

Subtract 500 from each side

500-500 = -4.9t^2 + 120t-500

0 =  -4.9t^2 + 120t-500

a=-4.9  b= 120 and c = -500

Using the quadratic formula

-b ±sqrt(b^2 -4ac)

2a

-120±sqrt(120^2 -4(-4.9)(-500))

2*(-4.9)

Solving this for the two solutions

t≈5.32415

t≈19.1656

The object will be above 500 meters between these two times

19.1656 - 5.32415 = 13.84145

This is approximately 14 seconds

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )

well my understanding of this question is they want you to use the y=14.1x -2.3 and put the coordinates they gave you onto the line segment of   y=14.1x -2.3

step-by-step explanation: