Answer:
[tex]1.5\; \rm s[/tex].
Step-by-step explanation:
The person in this question is in the air whenever the height is greater than [tex]0[/tex].
The graph of [tex]y = -16\, t^{2} + 24\, t[/tex] is a parabola opening downwards. Rearrange and find values of [tex]t[/tex] that would set this expression to [tex]0[/tex].
[tex]\begin{aligned}y &= -16\, t^2 + 24\, t \\ &= t\, (-16\, t + 24) = -8\, t\, (2\, t - 3)\end{aligned}[/tex].
The first factor, [tex]t[/tex], suggests that [tex]t = 0[/tex] would set this expression to [tex]0[/tex]- quite expected, since the person is on the ground right before jumping.
The second factor, [tex](2\, t - 3)[/tex], suggests that [tex]2\, t = 3[/tex] (in other words, [tex]t = 3/2 = 1.5[/tex]) would also set this expression to [tex]0[/tex]. Hence, this person would be once again on the ground [tex]1.5[/tex] seconds after jumping.
Hence, this person is in the air between [tex]t = 0[/tex] and [tex]t= 1.5[/tex] for a total of [tex]1.5\; \rm s[/tex].
