Time required for falling objects speed to double

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The discussion centers on calculating the time required for a dropped fish to double its speed after being released by an eagle flying horizontally at 7.6 m/s. Participants clarify that the question involves total speed, which combines both horizontal and vertical components. The vertical speed is determined by gravity, adding 9.8 m/s for each second. To find the total speed, the formula used is √(horizontal speed² + vertical speed²). Understanding these components is crucial for solving the problem accurately.
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An eagle is flying horizontally at 7.6 m/s with a fish in its claws. It accidentally drops the fish.
(a) How much time passes before the fish's speed doubles?
(b) How much additional time would be required for the fish's speed to double again?



Maybe someone else will understand this better than I am. There is no acceleration in the x direction, so they must be talking about the y direction. But if his initial y velocity was 0 (he is flying horizontally) then you can't double 0 ...



Thanks for any help.
 
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No, the question is talking about the total speed, both vertical and horizontal. Do you know how to find total speed given vertical & horizontal speed?
 
Is it √(x^2 + y^2)

I'm still not quite sure how to find the vertical component though?

Thanks for your reply.
 
Yes. The vertical component is just gt, since gravity adds 9.8 m/s to the vertical speed every second.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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