Let’s consider the vertical motion of a fog droplet subject to the force of gravity and drag. The droplet begins to move from rest. After a time, the droplet acquires a downward speed. For velocity v, the drag force is 6phr v upward. The acceleration of the droplet is determined by the equation
m a = mg – 6phr v
Dividing both sides by m we get
a = g – 6phr v / m
Thus the acceleration of the falling body depends on the instantaneous velocity. For v = 0, the initial acceleration is g. as the velocity increases, the two forces eventually become equal in magnitude. At this point, the net acceleration is zero; the velocity is at the maximum value and is called the terminal velocity. The terminal velocity of a falling object is the constant maximum velocity, that object attains and maintains in falling through a resistive medium. The fluid’s dragon the object is just sufficient to balance the weight of the object. Putting a = 0 in above equation we get.
0 = g – 6phr v / m
or 6phr v / m = g
6phr v = mg
v = mg / 6phr
This velocity is called terminal velocity so the above formula becomes
Vt = mg / 6phr
The terminal velocity is directly proportional to the object’s mass. The more massive an object, the faster it falls through a fluid. A cricket ball falls faster than a fog droplet.
TERMINAL VELOCITY OF PARATROOPERS: - The frictional resistance encountered by a man falling through the air increases with his speed until he reaches a terminal velocity, about 200km/hr, at which the retarding force of air resistance equals his weight. When the aviator opens the chute, the greater surface it presents increases the retarding force and reduces the terminal velocity, say 25km/hr, which is about equal to the velocity gained in jumping from height 2m. In such a descent the aviator reaches the ground with the same terminal velocity whether he jumps from an altitude of 200m or 2000m.
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