Question

A big cruise ship dropped anchor off the Caribbean island of Barbuda. The heavy anchor was released from a height of 54 meters above the water's surface, and it fell at a constant rate. After 35 seconds, the anchor was 9 meters below the water's surface. How fast did the anchor drop? How long did it take the anchor to reach the water's surface?

Answer 1

54+9= 63 m = distance and time=35 sec hence, speed= 63/35=9/5=1.8 m/s
time taken to reach water surface= distance/speed=54/1.8=30 sec

Answer 2

Answer:

Part 1) $1.8\frac{m}{s}$

Part 2) $30\ sec$

Step-by-step explanation:

Step 1

Find the speed at which the anchor dropped

we know that

The speed is equal to the distance divided by the time

In this problem the distance is equal to

$54+9=63\ m$

And the time is equal to $35\ sec$

substitute and find the speed

$\frac{63}{35}= 1.8\frac{m}{s}$

Step 2

Find the time taken for the anchor to reach the water's surface

Remember that

$speed=\frac{distance}{time}$

In this part we have

$speed=1.8\frac{m}{s}$

$distance=54\ m$

Substitute and solve for the time

$1.8=\frac{54}{time}$

$time=54/1.8=30\ sec$