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Problem 31 Hard Difficulty

SSM An eagle is flying horizontally at 6.0 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?

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Physics 101 Mechanics

Physics

Chapter 3

Kinematics in Two Dimensions

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Noor A.

October 2, 2021

I don't understand why the hypotenuse is 2×vx I understand that we're finding the time it takes to double in speed but why is it okay to just go ahead and multiply vx by 2 to get the hypotenuse. So basically I'm struggling with the concept.

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Video Transcript

So the question states that a bird flying horizontally at six meters per second drops of fish, and we need to find at what time the fish speed doubles. So to do this, we know that the horizontal velocity of fish is going to stay constant because it's project on motion. So it will always be six meters per second in the horizontal direction. Um, but the vertical velocity is going to change because it's accelerating downward. So to find when the magnitude of the velocity is 12 meters per second, we can draw components of, uh the fishes velocity. So we know that it's horizontal component is six meters per second. We don't know what the vertical bone it is and we know that the high pontin use the total magnitude is gonna be 12 meters per second. So we can solve for this using the question mark using possess ignorant there, Um, so 12 squared is equal to six squared plus question marks square, where the question mark is equal to 10 0.39 meters per second. So now that we know the vertical velocity of the fish, uh, and that will make it so that the total velocity of the fish. The magnitude of velocity is 12 meters per second. We can solve for the cinematics equation that says that the velocity at Time T is equal to the initial velocity plus a T. So the initial vertical velocity of the fish is zero. So this zero we know the final velocity of the fish to make them total magnitude of the velocity 12 meters second must be 10.39 meters per second. And we know the acceleration must be 9.8 meters per second squared so we can solve for the time and we find that T is equal to 1.1 seconds for the second half of the problem were asked to find how much late time later will it take for the fish to double again? So 2 24 meters per second and we basically use the same thing that we did for the first part of the problem. So we know the horizontal velocity of the fish is going to be six meters per second and we know we want this high pot news to be 24 meters per second. So we confined what the vertical velocity must be for this to be true. Once again, we're gonna have to use Pythagorean theorem. 24 squared is equal to six square plus question. Mark squared where question mark is equal to 23.23 meters per second. Once again, we can use the Kingdom axe formula, which states that the flossy a Time T is equal to that initial Aussie. Plus the acceleration times time in this case, our initial vertical velocity. We're gonna use the velocity from the first part of the problem. So 10.39 meters per second this because we're trying to find how much time after this point it will double. Um, the acceleration once again is just a 9.8 and our final velocity is just 23.23 And when we saw for tea, we find that it's equal to 1.31 seconds

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