An object is dropped from a small plane. As the object falls, its distance, d, above the ground after t seconds, is given by the formula d = –16t2 + 1,000. Use an inequatily equation to find for the interval of time taken by the object to reach the height greater than 300 feet above the ground
find where it equals 300
300=-16t²+1000
minus 1000 from both sides
-700=-16t²
divide both sides by -16
43.75=t²
square root both sides
6.614=t
it took at 6.614 seconds to get to a height of 300 ft above the ground
-16t^2 + 1000 > 300
16t^2 < 1000 - 300
16t^2 < 700
t^2 < 700/16
t^2 < 43.75
t < 6.61