Question

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An eagle is flying horizontally at 5.3 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 speed to double again?

Answer 1

The fish is initially moving horizontally at 5.3 m/s, with no vertical velocity.

When the eagle drops the fish, the fish is accelerated downward by gravity at 9.81 m/s^2

we know the final velocity for part a has to be 5.3 * 2 = 10.6 m/s
This is a vector, with components in the vertical and horizontal direction. Since we know the horizontal component remains constant, we can say

(10.6)^2 = (5.3)^2 + X^2

So X = 9.18 m/s is our vertical component of the velocity in order for the speed to double.

We know that final velocity (vertically) is v = v0 + at

a = gravity = 9.81 m/s^2

So
9.18 m/s = 0 + 9.81 * t

So t = 8.8334 / 9.81 = 0.936 seconds for the speed to double...... Answer.


Part B

To double again, the speed would have to be 21.2 m/s
20.1^2 = 5.3^2 + X^2
X = 20.53 m/s in vertical direction
v = v0 + at
20.53 = 0 + 9.81 * t
t = 2.09 seconds for the speed to double again.... Answer.

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