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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 t=0 112+96(0)-16 ( 0 ) 2 112+0-0=112 S(0)=112 When t=0, it’s means 0 second The ball ready be throwed. The building is 112ft. Now, the ball at 112ft. The building high is equal where the ball Thrown from. Now, the ball is going upward, and before the vertex. 0.5 =0.5 112+96(0.5)-16 ( 0.5 ) 2 112+48-4=156 S(0.5)=156 When t=0.5, the ball just thrown out, it going upward. So, the ball at 156ft at 0.5 sec. The ball is going upward, and before the vertex. 1 t=1 112+96(1)-16 ( 1 ) 2 112+96-16=192 S(1)=192 When t=1. The ball has been thrown out for 1second. Now, the ball at 192ft. The ball is going upward and before the vertex. 2 t=2 112+96(2)-16 ( 2 ) 2 112+192-64=240 S(2)=240 When t=2. It means the ball has been thrown out for 2 seconds. Now, the ball at 240ft. The ball is going upward and before the vertex. 100 = 112 + 96t – 16t 2 S(t)=112-100+96t-16

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