Question

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?

Answer 1

Answer:

8 seconds.

Step-by-step explanation:

To solve this question, we need to find the time, $t$, when $h = 0$, that is, when the height of the object above the ground is 0 feet.

Substituting $v = 128$ feet per second and $h = 0$ gives  the quadratic:

$0 = 128t - 16t^{2}$

Since 128 is divisible by 16, it can be reduced to $0 = 16(8t - t^{2})$.

We must now solve for $t$.

We can easily see that one answer to the equation is 0,

$8(0) - (0)^{2} = 0$ (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, $t = 8$ is also easy to see by inspection:

$8(8) - (8)^{2} = 0$.

Therefore the object lands 8 seconds after it is thrown.