Asmall fish is dropped by a pelican that is rising steadily at 0.500 m/s. how far below the pelican is the fish after 2.50 s?

Refer to the diagram shown below.

h = original height of the pelican when the fish is dropped (not relevant).
S =  distance traveled by the fish as a function of time, measured upward.
u = 0.5 m/s, the upward velocity with which the fish is dropped.
g = 9.8 m/s², the acceleration due to gravity.

Use the following equation:
S = ut + (1/2)gt²

S = (0.5 m/s)*(2.5 s) + 0.5*(-9.8 m/s²)*(2.5 s)²
   = -29.375 m

The negative sign means that the fish drops by  29.375 m from the original height of h.

The fish is 29.375 m below where the pelican dropped it after 2.5 s.

The best answer i can think of is b