An object is launched into the air from a ledge 6 feet off the ground at initial vertical velocity of 96 feet per second. Its height H, in feet, at t second is given by the equation H = 16t^2+96t+16. Find all the times t that the object is at a height of 160 feet off the ground.
+934 The question let's us set up the equation 160= 16t^2+96t+16 and we just solve for t to get t= 3(sqrt 2 - 1) and -3(sqrt 2 - 1).
Hope it helps!
