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?

(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