Question

A small fish is dropped by a pelican that is rising steadily at 0.50 m/s.

Answer 1

Answer:

(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