A cruise ship has decided to drop anchor off the Caribbean island of Antigua in 400 feet of water. The heavy anchor drops through the water to the sea floor at a constant rate. After dropping for 5 seconds the anchor is 360 feet above the sea floor and after dropping for 17 seconds the anchor is 264 feet above the sea floor. Once the anchor begins to drop, in how many seconds will the anchor reach the sea floor?

Question

anonymous · Answer

Hi, I wrote an  answer but it seems I misunderstood the question.
Anyway:
We know the anchor drops down at constant rate.
We know that 5 seconds 'after it started to drop' it was at height of $360_{ft}$ above the bottom.
And 17 seconds 'after it started to drop' (in here I thought 17 seconds after previous step...) it reached height of $264_{ft}$ above the bottom.
Means that it took it $17 - 5 = 12$ seconds for it to drop from height of $360_{ft}$ to height of $264_{ft}$, which means it dropped $ 360 - 264 = 96$ feet by that time.
So now, let's see how much did it drop per second:
$$
\frac{96_{ft}}{12_s} = 8_{ft/s}
$$
So the drop rate is therefore 8 feet per second.
So now let's found out how many steps like those are between the drop from height of $360_{ft}$ until it hits the ground.
The distance it has to pass is of course $360_{ft}$ so:
$$
\frac{360_{ft}}{8_{ft/s}} = 45_s
$$
So it took it 45 seconds to get from height to $360_{ft}$ to the bottom. Adding the 5 seconds it took it to get to that height:
$$
45_s + 5_s = 50_s
$$
So in total it took 50 seconds for it to drop from its initial position to the bottom.

The initial position of the anchor is the distance it passed in those 50 seconds so:
$$
50_s \times 8_{ft/s} = 400_{ft}
$$
Which means the anchor indeed was at the water level when started to drop.
Hope that helps