document.write( "Question 675752: if an object is propelled upward from a height of 64 feet at an initial velocity of 48 feet per second, then its height h after t seconds is given by the equation h=-16t^2+80t+48. after how many seconds did the object hit the ground \n" ); document.write( "
Algebra.Com's Answer #420035 by ewatrrr(24785)\"\" \"About 
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document.write( "Hi
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document.write( "h= -16t^2+80t+48 = 0   (object hitting the ground)
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document.write( " -16t^2+80t+48 = 0
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document.write( " 16t^2-80t-48 = 0
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document.write( " t^2-5t-3 = 0
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document.write( "\"t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
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document.write( "\"t+=+%285+%2B-+sqrt%28+37+%29%29%2F%282%29+\" ||Positive value only for unit measure
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document.write( "0r
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document.write( " -16(t-5/2)^2 + 100 + 48 = 0
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document.write( " -16(t-5/2)^2 + 148 = 0
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document.write( "     t = 5/2 ± sqrt(148/16)
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document.write( "     t = 5/2 ± sqrt(37)/2  \n" );
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