Question 597366: 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= -16tsquared + 64t + 6. Find all the times t that the object is at a height of 54 feet off the ground.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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= -16tsquared + 64t + 6.
Find all the times t that the object is at a height of 54 feet off the ground.
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Solve: -16t^2+64t+6 = 54
-16t^2+64t-48 = 0
-16(t^2-4t+3) = 0
t^2-4t+3 = 0
(t-3)(t-1) = 0
time = 1 second to get to 54 feet on the way up.
and t = 3 seconds to be at 54 feet on the way down.
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Cheers,
Stan H.
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