# Math 111 Module 4 - GD Using_Quadratic_Equation.docx -...

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Using a Quadratic Equation Problems A 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 t seconds is s (t) = 112 + 96t - 16t 2 Complete the table and discuss the interpretation of each point. t s(t) Interpretation 0 112 Because no time has elapsed the ball is still in its initial height of 112 ft s(t) = 112 + 96*0 – 16*0 2 s(t) = 112 0.5 156 If we substitute 0.5 as t then the ball moves upward 44 feet in the elapsed time. s(t) = 112 + 96*0.5 – 16*0.5 2 Simplify s(t) = 112 + 48 – 4 s(t) = 156 1 192 If we substitute 1.0 as t then the ball moves upward 80 feet in the elapsed time. s(t) = 112 + 96*1 – 16*1 2 Simplify s(t) = 112 + 96 – 16 s(t) = 192 2 240 If we substitute 2.0 as t then the ball moves upward 128 feet in the elapsed time. s(t) = 112 + 96*2 – 16*2 2 Simplify s(t) = 112 + 192 – 64 s(t) = 240
6.12 100 By substituting 100 into s(t) of the quadratic equation you get 6.12 seconds. Equation would be 100 – 112 + 96t + 16t 2 = 0 Simplified to -12 + 96t + 16t 2 = 0 Use quadratic equation as follows: 96 ± (− 96 ) ² 4 ( 16 12 ) 2 16