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The equation y=4.9t2+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
avatar+502 
0

Physics or Maths

 Dec 30, 2017
 #2
avatar+130466 
+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

 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+199 
+3

Setting y to zero, we find the following:

0=4.9t2+42t+18.9=49t2+420t+189=7t260t27=(7t+3)(t9)

As t must be positive, we can see that t=9.

 Dec 30, 2017

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