Explanation:
the ground = 0 ft.
so,
0 = -16t² + 40t + 6
let's simplify
0 = -8t² + 20t + 3
the general solution of a quadratic equation is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
x = t
a = -8
b = 20
c = 3
t = (-20 ± sqrt(400 - 4×-8×3))/(2×-8) =
= (-20 ± sqrt(400 + 96))/-16 =
= (-20 ± sqrt(496))/-16 = (-20 ± 4×sqrt(31))/-16
t1 = (-20 + 4×sqrt(31))/-16 = 20/16 - sqrt(31)/4 =
= 5/4 - sqrt(31)/4 = -0.141941091... s
t2 = (-20 - 4×sqrt(31))/-16 = 20/16 + sqrt(31)/4 =
= 5/4 + sqrt(31)/4 = 2.641941091... s
a negative time duration does not make sense for our scenario here, so t2 = 2.641941091... s is our solution here.
it takes the ball 2.641941091... seconds to hit the ground.