+1206 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?
+36915 When the ball hits the ground, y= 0 so substitute that in to the equation and solve for t
0 = -4.9t^2+42t +18.9 Use Quadratic Formula a = -4.9 b = 42 c = 18.9
\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
t=−0.428571 throw out
t=9 sec
+128406 The ball hits the ground when y = 0....so.....
-4.9t^2 + 42t + 18.9 = 0 multiply through by -10 to clear the decimals
49t^2 - 420t - 189 = 0 factor, if possible
(7t - 63) ( 7t + 3) = 0 set both factors to 0 and take the positive result
7t - 63 = 0 7t + 3 = 0
7t = 63 divide both sides by 7 7t = - 3
7 = 9 sec t = -3/7 sec (reject)
Here's a graph that verifies that the object hits the ground at 9 sec after launch
https://www.desmos.com/calculator/bppxhga5of
