# An eagle is flying horizontally at 5.2 m/s with a fish in its claws. It accidentally drops the fish.?

(a) How much time passes before the fish's speed quadruples?

(b) How much additional time would be required for the fish's speed to quadruple again?

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a

5.2 * 4 = 20.8

then

Vf = Vi + g t = 0 + 9.8 t =20.8

t = 2.1 s

b

20.8 *4 = 83.2

83.2 = 20.8 + 9.8 t

9.8 t = 62.4

t = 6.4 s

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• (a) Correct answer is not 2.1 s. Gravity acts downwards, it doesn't change the horizontal speed.

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• An eagle is flying horizontally at 5.2 m/s with a fish in its claws. It accidentally drops the fish.

(a) How much time passes before the fish's speed quadruples?

(b) How much additional time would be required for the fish's speed to quadruple again?

speed is the vector sum of 5.2 m/s horizontally, and the vertical speed which is

v = gt or t = v/g where g = 9.8 m/s²

the sum of the two = √(5.2² + (gt)²)

you want V₁ = 4•5.2 = 20.8 m/s

20.8 = √(5.2² + (gt)²)

5.2² + (gt)² = 432.64

(gt)² = 432.64 – 27.04 = 405.6

gt = 20.14

t₁ = 2.06 s

you want V₂ = 16•5.2 = 83.2 m/s

83.2 = √(5.2² + (gt)²)

5.2² + (gt)² = 6922.24

(gt)² = 6922.24 – 27.04 = 6895.2

gt = 83.04

t₂ = 8.47 s

t₂ – t₁ = 6.41 s

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• "Quadruple" means the speed goes from 5.2 to 20.8 m/s. If we ignore air resistance (these types of problems just about always ignore air resistance), then the the horizontal speed doesn't change until the fish hits something. So the downward speed needs to be 20.14 m/s.

20.14^2 + 5.2^2 = 20.8^2

With just gravity, how long does it take to get to a downward speed of 20.14 m/s?

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