Math 111 mod 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 s(t) Interpretation 0 112 s ( 0 ) = 112 + 96 ( 0 ) 16 ( 0 ) 2 S(0)=112 The ball is on the top of the building before it gets thrown. 0.5 156 0.5 ¿ 2 s ( 0.5 )= 112 + 96 ( 0.5 )− 16 ¿ S(0.5)=156 The ball is thrown vertically upwards and reaches a height of 156 feet. 1 192 1 ¿ 2 s ( 1 ) = 112 + 96 ( 1 ) 16 ¿ S(1)=192 The ball reaches a height of 192 feet when simultaneously is slowed by resistance forces 2 240 2 ¿ 2 s ( 2 ) = 112 + 96 ( 2 ) 16 ¿ S(2)=240 The ball reaches a height of 240 feet. t 1 =− 0.0 100 s ( t ) = 100 100 = 112 + 96 t 16 t 2 16 t 2 + 96 t + 2 = 0 (divide by -2) 8 t 2 48 t 1 = 0 t = 48 ± 48 2 4 ( 8 )(− 1 ) 2 ( 8 ) = 48 ± 2336 16 = 48 ± 4 146 16 = 12 146 4 =-0.02 Not possible. t 2 = 6.02 100 s ( t ) = 100 100 = 112 + 96 t 16 t 2 16 t 2 + 96 t + 2 = 0 8 t 2 48 t 1 = 0 (divide by -2) t = 48 ± 48 2 4 ( 8 )(− 1 ) 2 ( 8 ) = 48 ± 2336 16 = 48 ± 4 146 16 = 12 + 146 4 = 6.02 possible.
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