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A body of mass m is projected inside a liquid at an angle θ0 with horizontal at an initial velocity v0. If the liquid develops a velocity dependent force F= -kv where k is a positive constant, determine the x and y components of the velocity at any instant.

Solution:

A body of mass m is projected inside a liquid at an angle θ0 with horizontal at an initial velocity v0. The liquid develops a velocity dependent force F= -kv where k is a positive constant.

Hence,

m dv/dt= -kv

or, dv/dt= -k/m v

or, dv/v= -k/m dt

Integrating both sides,

∫dv/v = -k/m ∫dt

ln|v|= -kt/m+c (where c is a constant of integration)….. (i)

Now, for the x component of velocity,

ln vx=  -kt/m+ c

From the given problem, we have

vx=v0 cosθ0 at t=0

Applying the above condition in eqn.(i), we get

ln (vx/ v0 cosθ0 )= -kt/m

or vx=v0cosθ0e-kt/m

For the y component, we have to consider the acceleration due to gravity g.

Hence,

m dvy/dt= -kvy-mg

or, dvy/(kvy+mg)= -k/m dt

Integrating both sides,

ln|kvy+mg|=-kt/m+c

At t=0, vy=v0sinθ0

Hence,

kvy+mg=(kv0sinθ0+mg) e-kt