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
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Question 210198: 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 Alan3354(31529) (Show Source):
You can put this solution on YOUR website! H = -16t^2 + 64t + 6 = 54
-t^2 + 4t - 3 = 0
-(t-1)*(t-3) = 0
t = 1 second ascending
t = 3 seconds descending
