document.write( "Question 118531: A flare is fired as a stress signal and its height ,h in metres above the ground, t seconds after firing, is provided in the table below. Algebraically determine the quadratic function that defines the height of the flare above the ground t seconds after firing, and use it to determine the flare's height at 2.4 seconds.\r
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document.write( "t---0 1 2 3 4
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document.write( "h---2 97 182 257 322
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document.write( "I tried y=-2ak^2
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document.write( "97=-2a(1)^2
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document.write( "97/-2=-2a/-2
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document.write( "a=-48.5 and was completly lost :( Please Help Me :) \n" );
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Algebra.Com's Answer #86641 by scott8148(6628)![]() ![]() You can put this solution on YOUR website! the general equation is h=at^2+bt+c\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "by plugging in values from the table, you can find the values for a, b, and c\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "when t=0, h=c ___ so c=2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "when t=1, 97=a+b+2 ___ 95=a+b ___ 95-a=b\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "when t=2, 182=4a+2b+2 ___ 180=4a+2b ___ 90=2a+b\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "substitution gives 90=2a+(95-a) ___ -5=a ___ 95-(-5)=b ___ 100=b\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "h=-5t^2+100t+2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "in the real world, the a term is half the gravitational acceleration (negative because gravity is downward) \n" ); document.write( "___ the b term is the initial velocity of the flare \n" ); document.write( "___ and the c term is the initial height (probably the height of the person shooting the flare) \n" ); document.write( " |