# Module 4 Group Disscussion.docx - Using a Quadratic...

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Using a Quadratic Equation ProblemsA ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s(in feet) of the ball from the ground after tseconds is1.)s (t)= 112 + 96t - 16t221121129616tt96160ttt(9616 )0tt9616t6t2.) t = 0.5 s (t)= 112 + 96t - 16t2211296(0.5)16(0.5)s t211296(0.5)16(0.5)s t112484s t156s t3.) t = 1 s (t)= 112 + 96t - 16t2211296(1)16(1)s t211296(1)16(1)s t1129616s t192s t4.) t = 2 s (t)= 112 + 96t - 16t2211296(2)16(2)s t211296(2)16(2)s t11219264s t192s tStep 1.) Replace s(t) with 112 ft so we can solve for time (t).Step 2.) Subtract 112 from both sides of the equation allowing us t use the zero-product principle. Step 3.) Factor out a (t) Step 4.) t = 0 meaning at 0 seconds the ball is at its original height of 112ft Step 5.) Divide both sides by 16Step 6.) At 6 seconds the ball has returned to its original height of 112 ftStep 1.) Replace (t) with 0.5 ft so we can solve for distance s(t).Step 2.) Multiply 96 and by 0.5 and -16 by 2(0.5)Step 3.) Add and subtract the remaining numbersStep 4.) s(t) = 156 meaning at 005 seconds the ball is at a height of 156 ft Step 1.) Replace (t) with 1 ft so we can solve for distance s(t).Step 2.) Multiply 96 and by 1 and -16 by 2(1)Step 3.) Add and subtract the remaining numbersStep 4.) s(t) = 192 meaning at 1 seconds the ball is at a height of 192 ft Step 1.) Replace (t) with 2 ft so we can solve for distance s(t).Step 2.) Multiply 96 and by 2 and -16 by 2(2)Step 3.) Add and subtract the remaining numbersStep 4.) s(t) = 240 meaning at 2 seconds the ball is at a height of 240 ft
5 & 6.) s(t) = 100 s (t)= 112 + 96t - 16t221001129616tt210011296160tt21296160tt
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