math 111mod 4 discussion.docx - Using a Quadratic Equation...

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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 sec s(t) feet Interpretation 0 112 ft Set t=0 and solve 0 ¿ 2 = 112 s ( 0 ) = 112 + 96 ( 0 ) 16 ¿ s ( 0 ) = 112 At 0 seconds the ball will be at 112 ft 0.5 156 ft Set t=0.5 and solve 112+90(0.5)-16(0.5) 2 s ( 0.5 ) = 112 + 96 ( 0.5 ) 16 ( 0.5 ) 2 = 156 s ( 0 ) = 156 At 0.5 seconds the ball will be at 156 ft. 1 192 ft Set t=1 and solve 1 ¿ 2 = 196 s ( 1 ) = 112 + 90 ( 1 ) 16 ¿ s ( 1 ) = 192 At 1 second the ball will be 192 feet . 2 240 ft Set t=1 and solve 2 ¿ 2 = 240 s ( 2 ) = 112 + 90 ( 2 ) 16 ¿ s ( 2 ) = 240 At 2 seconds the ball will be 240 feet. 6.12 seconds 100 Use t = b 2 4 ac 2 a 96 ± 96 2 4 ( 16 )( 12 ) 2 (− 16 ) 96 ± 9984 32 96 9984 32 t ≈ 6.12 seconds 0.122 seconds 100 Using the quadratic formula to solve 100=112+96t-16t 2 ( 96 ) ±

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