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A ball is thrown into the air from the top of a building. The height, h(t), of the ball above the ground t seconds after it is thrown can be modeled by h(t) = -16t^2 + 64t + 80. How many seconds after being thrown will the ball hit the ground?

 Jan 28, 2020
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When the ball hits the ground    h will be = 0

 

h(t) = -16t^2 + 64t + 80

0 = h(t) = -16t^2 + 64t + 80      Solve for 't' using Quadratic Formula   

     One answer provided by the Quad Form will be -1    ......you want the POSITIVE  answer for 't'

 Jan 28, 2020

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