A small 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.

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