Question

Mark is in a deep hole looking for treasure. He is standing 8 feet below the surface. He throws an old coin he found with an initial upward velocity of 22 ft/sec. How long until it lands outside the hole, having gone up and come back down? Use the formula where h is the height of the coin in feet (relative to the surface) and t is the time in seconds since Mark threw it. Ignore air resistance and round your answer to the nearest tenth.

Answer 1

Answer:

Option D is correct.

Step-by-step explanation:

We have been given a function:

$0=-16t^2+22t-8$

We will find at h=0 to find the time it will take until it lands outside the hole, having gone up and come back down.

In the given function we will put h=0.

$-16t^2+22t-8=0$

On solving the above equation we will get t

$t=\frac{1}{16}(11-i\sqrt{7})\text{and}{\frac{1}{16}(11+i\sqrt{7})$

So, it will never happen we are getting time as imaginary value.

Hence, it will not make it outside the hole.

Therefore, option D is  correct.