Answer:
6.03s
Explanation:
Given the equation for the soccer ball's height, s, at time, t, seconds after the launch expressed as
s(t) = −4.9t2 + 19.6t +60
The ball hits the ground when s(t) = 0
Substitute s(t) = 0 into the expression
0= −4.9t2 + 19.6t +60
Multiply through by -10
0 = 49t²-196t-600
49t²-196t-600 = 0
Factorize
t = 196±√196²+4(600)(49)/2(49)
t = 196±√38416+117600/98
t = 196±√156016/98
t =196±394.99/98
t = 196+394.99/98
t = 590.99/98
t = 6.03secs
Hence the ball bit the ground after 6.03seconds later