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?