Answer:
8 seconds.
Step-by-step explanation:
To solve this question, we need to find the time, [tex]t[/tex], when [tex]h = 0[/tex], that is, when the height of the object above the ground is 0 feet.
Substituting [tex]v = 128 [/tex] feet per second and [tex]h = 0[/tex] gives the quadratic:
[tex]0 = 128t - 16t^{2}[/tex]
Since 128 is divisible by 16, it can be reduced to [tex]0 = 16(8t - t^{2})[/tex].
We must now solve for [tex]t[/tex].
We can easily see that one answer to the equation is 0,
[tex]8(0) - (0)^{2} = 0[/tex] (we need not concern ourselves with the 16 outside of the parenthesis as in the equation above, since 16 multiplied by 0 is 0). However this is the time the object is released into the air.
The second answer, [tex]t = 8[/tex] is also easy to see by inspection:
[tex]8(8) - (8)^{2} = 0[/tex].
Therefore the object lands 8 seconds after it is thrown.
