# 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

Algebra ->  Algebra  -> Quadratic Equations and Parabolas -> 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      Log On

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 Question 369536: 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 seconds is given by the equation H = -16t^2 + 64t + 6. Find all times t that the object is at a height of 54 feet off the ground.Answer by ewatrrr(10682)   (Show Source): You can put this solution on YOUR website! ``` 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 ```