Please answer fully :) 30 ptsAn 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

Answer:D) 14 secondsStep-by-step explanation:y = -4.9t^2 + 120tWe want to find out when y = 500500 = -4.9t^2 + 120tSubtract 500 from each side500-500 = -4.9t^2 + 120t-5000 =  -4.9t^2 + 120t-500a=-4.9  b= 120 and c = -500Using 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 solutionst≈5.32415 t≈19.1656 The object will be above 500 meters between these two times19.1656 - 5.32415 = 13.84145This is approximately 14 seconds