Asmall fish is dropped by a pelican that is rising steadily at 0.50 m/s. how far below the pelican is the fish after 2.5 s? a) 61 m b) 29.3 m c) 30.6 m d) 1.1 m can you tell me the formula in order to find this?

The pelican is rising at .5m/s. Therefore if the fish does not move at all, its distance below the pelican will increase at a rate of .50m/s. 
Let's now determine the distance from the starting point for the fish at time t.
 Downward acceleration due to gravity is 9.8m/(s^2).
 Initial downward velocity is -.50m/s because the pelican is initially holding the fish and rising.
 Velocity at time t is therefore = v0 + a * t = -.50m/s + (9.8m/(s^2)) * t
 Integrating velocity provides distance from initial position = v0 * t + (a / 2) * (t ^ 2)
 = -.50m/s * 2.5s + 4.9m/(s^2) * (2.5s)^2
 = 29.375m
 However, the pelican has moved .50m/s * 2.5s = 1.25m in the same time. Therefore the pelican is 1.25m above the initial position.
 Adding the fish's distance below the initial position and the pelican's distance above the initial position provides the fish's distance below the pelican:
 29.375m + 1.25m = 30.625m
 The answer is C.