Ask Question
12 November, 12:46

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?

+5
Answers (1)
  1. 12 November, 13:13
    0
    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.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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? ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers