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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?

 May 1, 2021
 #1
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When it hits the ground, y =   0...so....

 

We  just  need  to solve this

 

-4.9t^2  + 42t  +  18.9   = 0 

 

Using  the quad fromula  we get

 

[ -42 ± sqrt  (42^2  - 4 ( 18.9) ( -4.9) )]  / ( 2 * - 4.9)   

 

Taking the positive  value.....t  =  9   

 

( It  hits  the ground at 9 sec )

 

 

cool cool cool

 May 1, 2021

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