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 long did it take the anchor to reach the water's surface? How fast did the anchor drop?
Answers
Answer:
Answer:
Part 1) 1.8\frac{m}{s}1.8sm
Part 2) 30\ sec30 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\ m54+9=63 m
And the time is equal to 35\ sec35 sec
substitute and find the speed
\frac{63}{35}= 1.8\frac{m}{s}3563=1.8sm
Step 2
Find the time taken for the anchor to reach the water's surface
Remember that
speed=\frac{distance}{time}speed=timedistance
In this part we have
speed=1.8\frac{m}{s}speed=1.8sm
distance=54\ mdistance=54 m
Substitute and solve for the time
1.8=\frac{54}{time}1.8=time54
time=54/1.8=30\ sectime=54/1.8=30 sec
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Answer:
13.28m/s
Step-by-step explanation:
9 = u²/2g
∴ u = √(18g)
∴ u = 13.28 m/s