How Does Gravity Affect a Dropped Fish's Motion?

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The discussion centers on the motion of a fish dropped from a pelican ascending at 0.50 m/s. After 2.5 seconds, the fish's velocity is calculated to be -24 m/s, indicating downward motion due to gravity. The distance below the pelican after the same time interval is determined to be 31 meters. The participant correctly applies the equations of motion, specifically Vf = Vi + (A)(T) and ΔX = (Vi)(t) + (1/2)(A)(T)^2, to arrive at these conclusions.

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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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