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The equation \(y = -4.9t^2 + 42t + 18.9\) describes the height (in meters) of a ball tossed up in the air at 42 meters per second from a height of 18.9 meters from the ground, as a function of time in seconds. In how many seconds will the ball hit the ground?

 Dec 2, 2020
 #1
avatar+114592 
+1

When it hits the gtound the height will  = 0

 

So

 

-4.9t^2  + 42t +  18.9  = 0

 

Using the quadratic formula we  have  that

 

t =   -42 ±  sqrt ( 42^2 - 4(-4.9)(18.9) )           -42 ±  46.2          -42 - 46.2

     ____________________________  =    ___________  =  ________   =   9 sec

                     2 (-4.9)                                          -9.8                   - 9.8 

 

 

cool cool cool                                  

 Dec 2, 2020
 #2
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+1

Thanks!

 Dec 2, 2020

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