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Due to air a falling body faces a resistive force proportional to square of velocity \( v, \) consequently its effective downward acceleration is reduced and is given by \( a=g-k v^{2} \) where \( k= \) \( 0.002 m^{-1} . \) The terminal velocity of the falling body is \( (\operatorname{in} \mathrm{m} / \mathrm{s}) \)

1. 60
2. 70
3. 80
4. 90

Correct answer: 80

Solution

Terminal velocity occurs when downward acceleration is exactly balanced by air resistance, so the effective acceleration is zero. Using a = g - kv^2 and setting a = 0 gives v = sqrt(g/k), which evaluates to 80 m/s.

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