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5 July, 02:20

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?

b. How far below the pelican is the fish after 2.5 s?

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  1. 5 July, 02:22
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    (a) 24.025 m/s. downward.

    (b) 31 m

    Explanation:

    From Newton's equation of motion,

    (a)

    v = u + gt ... Equation 1

    Where v = final velocity, u = initial velocity, g = acceleration due to gravity, t = time.

    Note: Let upward velocity be negative and downward be positive

    Given: u = - 0.5 m/s (upward), t = 2.5 s

    Constant : g = 9.81 m/s²

    Substitute into equation 1

    v = 0.5+9.81 (2.5)

    v = - 0.5+24.525

    v = 24.025 m/s. downward.

    (b) using

    s₁ = ut + 1/2gt² ... Equation 2

    Where s₁ = distance at which the fish fall after being dropped by the pelican

    Given: u = - 0.5 m/s, t = 2.5 s, g = 9.81 m/s²

    Substitute into equation 2

    s₁ = - 0.5 (2.5) + 1/2 (9.81) (2.5) ²

    s₁ = - 1.25+30.656

    s₁ = 29.41 m

    also,

    s₂ = vt ... Equation 3

    Where s₂ = the distance by which the pelican rise during this time.

    Given: v = 0.5 m/s, t = 2.5 s

    s₂ = 0.5 (2.5)

    s₂ = 1.25 m.

    Note: Distance between the pelican and fish = s₁ + s₂

    Distance between the pelican and fish = 29.41+1.25

    Distance between the pelican and fish = 30.66

    Distance between the pelican and fish ≈ 31 m
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