Question

A small mailbag is released from a helicopter that is descending steadily at 1.07 m/s.

(a) After 5.00 s, what is the speed of the mailbag?

v = _____ m/s

(b) How far is it below the helicopter?

d = _____ m

(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.07 m/s?

v = _____ m/s

d = _____ m

Answer 1

Let us assume that downward direction is positive and upwards direction as negative

Velocity of helicopter = V_h = 1.07 m/s (remains constant)

Initial velocity of mail = V₀ = 1.07 m/s

Acceleration of mail = g = 9.81 m/s²

Time = t = 5 sec

a) Speed of mail after 5 sec = V

V = V₀ + gt

V = 1.07 + 9.81 x 5

V = 50.12 m/s

b) Distance of mail from helicopter after 5 sec = d

Distance travelled by mail in 5 sec = s₁

Distance travelled by helicopter in 5 sec = s₂

s₁ = V₀t + (gt²)/2

s₁ = 1.07 x 5 + (9.81 x (5)²)/2

s₁ = 127.975 m

s₂ = V_ht + (at²)/2

Here acceleration = 0 m/s² as helicopter is moving steadily, hence maintains constant velocity.

s₂ = 1.07 x 5 + ( 0x(5)²)/2

s₂ = 5.35m

d = s₁ - s₂

d = 127.975 - 5.35

d = 122.625 m

c) If the helicopter was rising steadily at 1.07 m/s

Velocity of helicopter = V_h = -1.07 m/s (remains constant)

Initial velocity of mail = V₀ = -1.07 m/s

Acceleration of mail = g = 9.81 m/s²

Time = t = 5 sec

Speed of mail after 5 sec = V

V = V₀ + gt

V = -1.07 + 9.81 x 5

V = 47.98 m/s

Distance of mail from helicopter after 5 sec = d

Distance travelled by mail in 5 sec = s₁

Distance travelled by helicopter in 5 sec = s₂

s₁ = V₀t + (gt²)/2

s₁ = -1.07 x 5 + (9.81 x (5)²)/2

s₁ = 117.275 m

s₂ = V_ht + (at²)/2

Here acceleration = 0 m/s² as helicopter is moving steadily, hence maintains constant velocity.

s₂ = -1.07 x 5 + ( 0x(5)²)/2

s₂ = -5.35m

d = s₁ - s₂

d = 117.275 - (-5.35)

d = 122.625 m