Question

Mark is in a deep hole looking for treasure. He is standing 18 feet below the surface. He throws an old ring he found with an initial upward velocity of 33 ft/sec. How long until it lands outside the hole, having gone up and come back down? Use the formula h= -16t^2+33t-18, where h is the height of the ring 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.

A. 0.4 seconds
B. 1.9 seconds
C. 2.5 seconds
D. It will not make it outside the hole.

Answer 1

D. It will not make it outside the hole

Answer 2

Answer:

It will not make it outside the hole

Explanation:

It is given that,

Mark is in a deep hole looking for treasure. He is standing 18 feet below the surface.  He throws an old ring he found with an initial upward velocity of 33 ft/sec.

Using the formula, $h=-16t^2+33t-18$........(1)

Where

h is the height of the ring in feet, t is in seconds since Mark threw it.

On solving quadratic equation (1) as :

$-16t^2+33t-18=0$

Using the relation :$t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$t=\dfrac{-33\pm\sqrt{33^2-4(-16)(-18)}}{2(-16)}$

The solution (time) of above equation is complex. So, the correct option is (D) " It will not make it outside the hole"