can anyone pls solve this for me?a rain drop falls from rest in an atmosphere saturated with water vapour. as it falls, water vapour condenses on the drop at the rate of mass mu per second. if initial mass of drop is m naught, how much distance the drop falls in time t?
62Lets assume that the mass at any time is m(t)
and velocity is v(t)
mass at time t+dt = m+dm
Velocity at time t+dt = v+dv
External impulse during the time is (m)g dt
The change in momentum is impulse
(m+dm)(v+dv)-mv=mgdt
mdv+vdm=mgdt
m(dv/dt)+v(dm/dt)=mg
m(dv/dt)=mg-vμ
Now solve the differential equation. from time zero to time t
1vr = v, dmdt=μ , Ft ( thrust ) = vrdmdt = vμ
Net force ( Fnet ) = F - Ft
→ m dvdt = F - μv
→ (m0 + μt )dvdt = F - μv
→\int_{0}^{v}{\frac{dv}{F-\mu v}} = \int_{0}^{}t{\frac{dt}{m_{0}+\mu t}}
which gives → v = Ftm0+μt
now use equation v2 - u2 = 2aS
since u = 0
(Ftm0+μt)2 = 2gS
which gives required S.