document.write( "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. \n" ); document.write( "

Algebra.Com's Answer #263342 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi,
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document.write( "H = -16t^2 + 64t + 6
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document.write( "Find all times t that the object is at a height of 54 feet off the ground.
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document.write( "54 = -16t^2 + 64t + 6
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document.write( "16t^2 - 64t -6 + 54 = 0
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document.write( "16t^2 - 64t + 48 = 0
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document.write( "16[t^2 - 4t + 3] = 0
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document.write( " t^2 - 4t + 3 = 0
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document.write( "factoring
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document.write( "(t-3)(t-1) = 0
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document.write( "(t-3)= 0  t = 3 sec
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document.write( "(t-1)= 0  t = 1 sec\r
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