Question 1162208: Daniel does a cannonball into the water. He hits the water with a speed of 12.0 m/s. The buoyant force of the water causes a constant acceleration of 8.00 (m/s)/s. This slows him down until he stops. Then, it pushes him back to the surface. How long is he under the water?
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Daniel does a cannonball into the water. He hits the water with a speed of 12.0 m/s. The buoyant force of the water causes a constant acceleration of 8.00 (m/s)/s. This slows him down until he stops. Then, it pushes him back to the surface. How long is he under the water?
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He descends for 12/8 = 1.5 seconds
The average speed during descent is 12/2 = 6 m/sec
6*1.5 = 9 meters below the surface at his greatest depth
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Then he accelerates upward for 9 meters.
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t = 1.5 seconds
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Total time submerged = 3 seconds
Answer by ikleyn(52748)  (Show Source):
You can put this solution on YOUR website! .
This problem, as it is formulated in the post, is absolutely non-physical,
and I will explain it to you right now, why it is so.
It is true that for the human body, the buoyant force is close the the weight.
The weight makes free fall acceleration of about 9.8 m/s^2;
the buoyant force in this problem makes deceleration of 8 m/s^2,
and it is close to be true.
But if even the buoyant force would be equal to the weight, the body would be still continue
to move vertically down in the water uniformly at the speed of 12.0 m/s.
But in reality, it does not happen.
Why ?
Because there is a THIRD FORCE, which is not accounted in the problem.
This force is DRAG of the water.
It produces the major deceleration to the falling body, in this case.
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The resume is that this problem has no right to exist,
since it contradicts to both common sense and observed facts.
It, probably, does not matter to you, who created this problem,
but in this case I care about next generation of students, who, probably, will read this post in the future . . .
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