Physics · Viscosity, Stoke's law, terminal velocity, streamline and turbulent flow, critical velocity
Due to air a falling body faces a resistive force proportional to square of velo
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 \( (\operatorname{in} \mathrm{m} / \mathrm{s}) \)
- A. 60
- B. 70
- C. 80
- D. 90
Step-by-step solution
Terminal velocity occurs when acceleration a = 0, so 0 = g - k v_t^2 => v_t = sqrt(g/k). Given g = 9.8 m/s^2 and k = 0.002 m^{-1}, v_t = sqrt(9.8/0.002) = sqrt(4900) = 70 m/s.
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