An object falling from rest in air is subject notonly to the gravitational force but also to air resistance. Assume that theair resistance is proportional to the velocity with constant ofproportionality as `k >0`, and acts in a direction opposite to motion `(g=9. 8 m/(s^2))dot`Thenvelocity cannot exceed.(c)`( d ) (e) 9.8//k""m//s (f)`(g)(b) `( h ) (i) 98//k""m//s (j)`(k)(c)`( d ) (e) (f) k/( g )(( h ) 9.8)( i ) (j)m//s (k)`(l) (d) None of theseA. 9.8/km/sB. 98/k m/sC. `k/9.8 m//x`D. None of these

An object falling from rest in air is subject notonly to the gravitati

1.	An object falling from rest in air is subject notonly to the gravitational force but also to air resistance. Assume that theair resistance is proportional to the velocity with constant ofproportionality as `k >0`, and acts in a direction opposite to motion `(g=9. 8 m/(s^2))dot`Thenvelocity cannot exceed.(c)`( d ) (e) 9.8//k""m//s (f)`(g)(b) `( h ) (i) 98//k""m//s (j)`(k)(c)`( d ) (e) (f) k/( g )(( h ) 9.8)( i ) (j)m//s (k)`(l) (d) None of theseA. 9.8/km/sB. 98/k m/sC. `k/9.8 m//x`D. None of these
Answer» Correct Answer - A Let V(t) be the velocity of the object at time t. Given `(dV)/(dt) =9.8-kV` or `(dvV)/(9.8-kV)=dt` Integrating, we get `log(9.8-kV) =-kt+logC` or `9.8-kV=Ce^(-kt)` But `V(0)=0` or `C=9.8` Thus, `9.8 - kV=9.8e^(-kt)` or `V(t)=9.8/k(1-e^(-kt)) lt 9.8/k` for all t Hence, V(t) cannot exceed 9.8k m/s.

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