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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?

tertre  Dec 30, 2017
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4+0 Answers

 #1
avatar+502 
0

Physics or Maths

Rauhan  Dec 30, 2017
 #2
avatar+80903 
+2

When it hits the ground, y  will   = 0

 

So we have

 

-4.9t^2  + 42t  +  18.9  =  0       multiply through by -10

 

49t^2  -  420t - 189  =  0           divide through by 7

 

7t^2  -  60t  - 27 =  0        factor

 

(7t  + 3 )  ( t  -  9 )  = 0

 

Setting  each factor to 0 and solving for t   gives

 

t =  - 3/7    sec      { reject }

And

t  =  9  sec

 

 

cool cool cool

CPhill  Dec 30, 2017
 #3
avatar+502 
0

I had my doubts that it was related to physics but didn’t bother doing it

Rauhan  Dec 30, 2017
 #4
avatar+64 
+2

Setting \(y\) to zero, we find the following:

\(\begin{align*} 0& = -4.9t^2 + 42t + 18.9\\ & = -49t^2 + 420t + 189\\ & = 7t^2 - 60t - 27\\ & = (7t + 3)(t - 9) \end{align*}\)

As \(t\) must be positive, we can see that \(t = \boxed{9}.\)

azsun  Dec 30, 2017

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