SOLUTION: An object is launched straight up into the air at an initial velocity of 64 feet per second. It is launched from a height of 6 feet off the ground. Its height H, in feet, at t seco

Hi,

H = -16t^2 + 64t + 6

Find all times t that the object is at a height of 54 feet off the ground.

54 = -16t^2 + 64t + 6

16t^2 - 64t -6 + 54 = 0

16t^2 - 64t + 48 = 0

16[t^2 - 4t + 3] = 0

 t^2 - 4t + 3 = 0

factoring

(t-3)(t-1) = 0

(t-3)= 0  t = 3 sec

(t-1)= 0  t = 1 sec