Gilstrap, J. M4 Quadratic Equation.docx - Using a Quadratic...

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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
t s(t) Interpretation 0 =112+96t-16t 2 112=112-96t-16t 2 96t-16t 2 =0 t(96-16t)=0 96=16t T=6 Replace s(t) with 112 ft, solve for time (t) Subtract 112 from both sides allowing the use of zero-product principle Factor out (t) t = 0 meaning at 0 seconds the ball is at original height of 112ft Divide both sides by 16 At 6 seconds the ball has returned to its original height of 112 ft 0.5 s(t)=112+96(.5)-16(.5) 2 s(t)=112+48-4 s(t)=156 Replace (t) with .5 ft to solve for distance Multiply 96/.5 and 16/(.5) 2 Add and subtract (112+48-4) At a half a second the ball is at 156 ft high 1 s(t)=112+96(1)-16(1) 2 s(t)=112+96-16 s(t)=192 Replace (t) with 1 ft to solve for distance Multiply 96/1 and 16/(1) 2 Add and subtract (112+96-16) At one second the ball is at 192 ft high 2 s(t)=112+96(2)-16(2) 2 s(t)=112+192-64 s(t)=240 Replace (t) with 2 ft to solve for distance

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