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A ball is thrown from the top of a building. The height, h-meters above ground, can be moddled by h = 12 - 5t - 2t^2. where t is the time after it leaves the ground

 

A. What is the height of the ball before it touches the ground?

 

B.At what time will the ball strike the ground again?

Guest Aug 19, 2018
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A  I assume this question means: " what is the initial height of the ball?"    Let t= 0

Then h = 12  - 5(0)- 2(0)^2 = 12 meters

 

B. This occurs when h= 0 

  0 = 12 - 5t - 2 t^2         

0 = 2t^2 +5t - 12

  Use Quadratic Formula to find    t = 1 1/2 sec

Guest Aug 19, 2018

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