Question

2) A fire-fighter helicopter carries a large bucket of water at the end of a cable. The speed of the helicopter is 65 m/s and air resistance results in a cable angle of 40°. 'What would be the terminal velocity of the bucket if dropped? (Assume that the resistive force is described by R = bv 2.) DASAL ob

Answer 1

The free-body diagram for the bucket before it is dropped is

mg

Where

F is the resistive force due to air, T is the tension in the cable, m is mass of the bucket and g is the acceleration due to gravity.

The resistive force is

F = bu?

Where v is the velocity of the helicopter.

At equilibrium

T cos = mg

T= mg cos

and

T sin0 = F = bu?

mg sin = bu? cos A

b= mgtand

Now, when the bucket is dropped, it reaches the terminal velocity when the net force on it becomes zero. i.e

mg - bu = 0

mg= but

6u = 11 €

Putting the value of b

Ut = 1mg matan

Ut= V tan

Putting the values

V = 652 Ut = V tan 40° 6 = 70.96 m/s

So, the terminal velocity is 70.96 m/s.