+0

# help sorry to bother...

0
185
3
+1195

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?

Jul 28, 2019

#1
+19931
0

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

Jul 28, 2019
#2
+106536
+1

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

Jul 28, 2019
#3
+1195
+1

Thank you ElectricPavlov and CHphill you guys are the best!!

Jul 28, 2019