Answer:
0.94secs
Explanation:
Given the function below that models the distance of the softball from the ground in meters after t seconds expressed as;
h(t)=-4.9t^2+5t+9
The ball hits the ground when the height h(t) is zero
Substitute h(t) = 0 into the expression and calculate the value of t;
0 = -4.9t²+5t+9
Multiply through by 10;
0 = 49t²+50t+90
-49t²+50t+90 = 0
Factorize;
t = -50 ±√50²-4(49)(90)/2(49)
t = -50±√2500+17640/98
t = -50±√20140/98
t = -50+141.92/98
t = 91.92/98
t = 0.94 secs
Hence it takes 0.94secs for the softball to hit the ground