A seagull flying horizontally over the ocean at a constant speed of 2.85 m/s carries a small fish in its mouth. It accidentally lets go of the fish, and 2.45 s after the bird lets go, the fish lands in the ocean. a. Just before reaching the ocean, what is the horizontal component of the fish's velocity?
b. Just before reaching the ocean, what is the vertical component of the fish's velocity?
c. If the seagull's initial speed were decreased, which of the following regarding the fish's velocity upon reaching the ocean would be true?

1. The horizontal component of the fish's velocity would decrease.
2. The horizontal component of the fish's velocity would increase.
3. The horizontal component of the fish's velocity would stay the same.
4. The vertical component of the fish's velocity would decrease.
5. The vertical component of the fish's velocity would increase.
6. The vertical component of the fish's velocity would stay the same.

a) 2.85 m/s

b) -24.01 m/s

c) 1

Explanation:

From the question, we can attest that the motion of the fish that was dropped by the seagul is that of a projectile motion. This motion is made up of two other motions which are, a horizontal uniform motion and a vertical motion, at constant acceleration. From the question, we are asked to find the horizontal motion. And then, the horizontal component of the fish's velocity does not change, therefore its the horizontal component of the fish's velocity is 2.85 m/s

To find the vertical component of the fish's velocity, we use the equation

v(y) = u(y) + gt

Where u(y) is the initial velocity which is zero, and g is the acceleration due to gravity which is -9.8 m/s. We are also told from the question that it took 2.45 s, and that's our time t. Applying this into the equation, we have

v(y) = 0 + -9.8 * 2.45

v(y) = -24.01 m/s

For the 3rd part, the horizontal component of the fish's velocity would decrease

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