Question

A seagull flying horizontally over the ocean at a constant speed of 3.15 m/s carries a...

A seagull flying horizontally over the ocean at a constant speed of 3.15 m/s carries a small fish in its mouth. It accidentally lets go of the fish, and 1.85 s after letting go the fish lands in the ocean.

(a) Just before reaching the ocean, what is the horizontal component of the fish's velocity? Ignore air resistance. Assume the bird is initially traveling in the positive x direction. (Indicate the direction with the sign of your answer.)

__________ m/s

(b) Just before reaching the ocean, what is the vertical component of the fish's velocity? Ignore air resistance. Assume upward is the positive y direction and downward is the negative y direction. (Indicate the direction with the sign of your answer.)

__________ m/s

(c) If the seagull's initial speed were increased, which of the following regarding the fish's velocity upon reaching the ocean would be true? (Select all that apply.)

The horizontal component of the fish's velocity would increase.

The horizontal component of the fish's velocity would decrease.

The horizontal component of the fish's velocity would stay the same.

The vertical component of the fish's velocity would increase.

The vertical component of the fish's velocity would decrease.

The vertical component of the fish's velocity would stay the same.

Homework Answers

Answer #1

a)

consider the motion of the fish after let go in horizontal direction or X-direction

Vox = initial velocity = 3.15 m/s

ax = acceleration along the horizontal direction = 0 m/s2

t = time taken = 1.85 s

Vfx = final velocity of fish just before hitting the ocean

using the equation

Vfx = Vox + ax t

Vfx = 3.15 + 0 (1.85)

Vfx = 3.15 m/s

b)

consider the motion of the fish after let go in horizontal direction or Y-direction

Voy = initial velocity = 0 m/s

ay = acceleration along the verticaldirection = - 9.8 m/s2

t = time taken = 1.85 s

Vfy = final velocity of fish just before hitting the ocean

using the equation

Vfy = Voy + ay t

Vfy = 0 + (- 9.8) (1.85)

Vfy = - 18.13 m/s

c)

The horizontal component of the fish's velocity would increase.

The vertical component of the fish's velocity would stay the same.

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