Answer:
Approximately [tex]9.31\; {\rm s}[/tex].
Step-by-step explanation:
Observe that initially when [tex]t = 0[/tex], the height [tex]h[/tex] of the object according to the given equation would be [tex]1\, 200[/tex] feet- same as the height of the waterfall relative to the ground. Thus, the height in this question are measured relative to the ground, and object would hit the ground when its height becomes [tex]0[/tex].
Therefore, to find the time when the object would hit the ground, set height [tex]h[/tex] to [tex]0[/tex] and solve for time [tex]t[/tex] ([tex]t \ge 0[/tex]):
[tex](-16)\, t^{2} + 20\, t + 1200 = 0[/tex].
Using the quadratic formula, the two roots of this quadratic equation would be:
[tex]\begin{aligned}t &= \frac{-(20) \pm \sqrt{(20)^{2} - 4\, (-16)\, (1\, 200)}}{2\, (-16)}\end{aligned}[/tex].
Either [tex]t \approx (-8.06)[/tex] or [tex]t \approx 9.31[/tex]. However, since t represents the time after launch, the value of [tex]t[/tex] must be non-negative ([tex]t \ge 0[/tex]) and [tex]t \approx 9.31[/tex] would be the only valid solution.
In other words, the object would hit the ground approximately [tex]9.31\; {\rm s}[/tex] after launch.
