Kinematics in 2d of eagle flying

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SUMMARY

The discussion focuses on the kinematics of an eagle dropping a fish while flying horizontally at 6.0 m/s. The key calculations involve determining the time it takes for the fish's speed to double from 6.0 m/s to 12.0 m/s, which is established as 1.1 seconds, and then to double again to 24.0 m/s, requiring an additional 1.3 seconds. Participants clarify that the fish's initial horizontal speed remains constant while it accelerates vertically due to gravity, necessitating the use of the Pythagorean theorem to find the resultant speed. The importance of correctly interpreting initial conditions and the role of gravity in kinematic equations is emphasized.

PREREQUISITES
  • Understanding of basic kinematics principles
  • Familiarity with the Pythagorean theorem
  • Knowledge of horizontal and vertical velocity components
  • Concept of gravitational acceleration (9.81 m/s²)
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  • Study the equations of motion under constant acceleration
  • Learn how to apply the Pythagorean theorem in two-dimensional motion problems
  • Explore the effects of air resistance on falling objects
  • Investigate real-world applications of kinematics in wildlife behavior
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1. An eagle is flying horizontally at 6.0m/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?



2. Maybe I'm thinking too hard about this problem, but at this moment i truly have no idea how to do this. It just seems to not make sense to me because wouldn't the amount of time it takes to double depend on the initial speed it gives you? I have the answers in the back of the back and the answers it gives me are (a) 1.1s (b) 1.3s



3. Alright well this chapter was labeled kinematics in 2d so i tried finding the velocity of the fish, and since i was not given the initial time, i subbed in 1. The horizontal component given was 6.0 m/s and the vertical component is the acceleration due to gravity. To find the combined component, i used the Pythagorean theorem and got ~11.49 m/s. This answer means to me that at 1 second, the speed of the fish is 11.49m/s. To find twice the speed, i multiplied it by 2 and squared it, getting ~528. I replugged that into the Pythagorean theorem making it 528 = 6^2(no acceleration) and (10t)^2. Solving this i got 2.22 s...which is not the correct answer.

Was there enough information given in the problem to solve it? Any help would be greatly appreciated =P
 
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I don't understand why you have assumed the time would be 1 initially. Take the time the eagle drops the fish as 0. You're along the right lines with the method however. When the fish is first dropped at t=0 it is traveling at 6m/s horizontally. Now you have to find out what the time will be when the fish is traveling at 12m/s.

That should be enough to get you the right answer I hope.
 
The fish has an initial speed of 6 m/s (the velocity is horizontal and let's assume constant if we neglect air resistance). Remember speed is simply the magnitude of velocity without regard to direction.

When the fish is dropped, it starts accelerating due to gravity, so one has to determine the vertical velocity, and then add the horizontal and vertical velocities, such that the resultant speed is 2 x 6 m/sec = 12 m/s.

then the speed again doubles to 24 m/s, but still with the same horizontal speed.

the Pythagorean theorem is correct.
 
I was thinking in terms of velocity so comparing the horizontal velocity and the combined velocity just didn't make sense to me. I feel kind of stupid now haha. Thanks for the help though :smile:
 

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