An object has been thrown straight up into the air. The formula h = vt - 16t^2 gives the height of the object above the ground after t seconds, when it is thrown upward with an initial velocity v. After how many seconds will the object hit the ground if it is thrown with a velocity of 128 feet per second?

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.