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# help!

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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 30, 2017

#1
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Physics or Maths

Dec 30, 2017
#2
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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   Dec 30, 2017
#3
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I had my doubts that it was related to physics but didn’t bother doing it

Rauhan  Dec 30, 2017
#4
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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 $$t = \boxed{9}.$$

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Dec 30, 2017