An object falling from rest in air is subject not only to the gravitational force but also to air resistance. Assume that the air resistance is proportional to the velocity with constant of proportionality as `k >0` , and acts in a direction opposite to motion `(g=9. 8 m/(s^2))dot` Then velocity 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 these

Options

A. 9.8/km/s
B. 98/k m/s
C. `k/9.8 m//x`
D. None of these

Answer

A

Explanation

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.