How Does Gravity Affect a Dropped Fish's Motion?

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A small fish dropped by a pelican ascending at 0.50 m/s experiences gravitational acceleration of -9.81 m/s². After 2.5 seconds, the fish's velocity is calculated to be -24 m/s, indicating it is moving downward. The distance below the pelican after the same time is determined to be 31 meters. The initial velocity of the fish is equal to the pelican's velocity at the moment of release. The user confirms they have resolved their question regarding the distance calculation.
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Question:

54) A small fish is dropped by a pelican that is rising steadily at 0.50 m/s.

a) After 2.5 s, what is the velocity of the fish? Ans: -24 m/s

b) How far below the pelican is the fish after 2.5 s? Ans: 31m

My Work/Question:

The fish would be moving at the same velocity as the pelican, which equals the (Vi). The Accel would equal -9.81 m/s^2 due to gravity.

I used Vf = (Vi) + (A)(T) to find the velocity of the fish.

My question on B: Would I use the... delt X = (Vi)(t) + (1/2)(A)(T)^2...formula for B?
 
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