Liam asked in Science & MathematicsPhysics · 9 months ago

# 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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• 9 months ago

...

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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• 9 months ago

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