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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

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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 8, 2020

#1
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Quadratic formula for ax^2 + bx + c: x = (-b pm sqrt(b^2 - 4ac))/2a

Plug into the formula to get t = 23 seconds

Dec 8, 2020
#2
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0 ht  is when the ball hits the ground

0 = -4.9 t^2 + 42t  + 18.9      a = -4.9     b = 42     c = 18.9    for Quadratic Formula:

$$t = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

Sub in the values to find    t = 9 seconds     or     - .428 (throw out )

Dec 8, 2020