When a ball is thrown upward, its height, h, in metres, is given by h=1.5 +19.6t -4.9t², where t is the number of seconds after it is thrown. For what length of time is the ball above the ground?

Answer 1

the ball is above the ground when h > 0

1.5 + 19.6t - 4.9t^2 > 0

49t^2 - 196t - 15 < 0
let's solve it as an equation.....
t = (196 ± √41356)/98 = 4.075 or a negative t
but t ≥ 0

so the ball is above the ground between
t=0 and t=4.075
that is, it was above ground for 4.075 seconds