Mathematics, 30.10.2019 20:31, breahnareedy
Aball is thrown upward from the top of a building. the ball's height above the ground after t seconds is given by the function: h(t) = -16t^2+48t+32.
a. what is the initial height (i. e. the height of the building)?
b. how high did the ball go?
c. when does the ball hit the ground?
Answers: 3
. For  this we can use the first derivative rule.
![t\[tex] such that [tex]h'(t)=0](/tpl/images/0353/1218/19219.png)
, this is 
. Then, evaluate the function 
in 
![a=-16,\ b=48,\ c=32[\tex] and taking the positive solution.[tex]x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-(48) \pm \sqrt{b(48)^2-4(-16)(32)}}{2(-16)} = \frac{-48\pm \sqrt{2304+2048}}{-32}\\\\=\frac{-48+\sqrt{4352}}{-32}\approx 3.56](/tpl/images/0353/1218/6c22a.png)
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Aball is thrown upward from the top of a building. the ball's height above the ground after t second...
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