- #1
Inonin
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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
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