A body of mass m is projected inside a liquid at an angle θ_{0} with horizontal at an initial velocity v_{0}. 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 θ_{0} with horizontal at an initial velocity v_{0}. 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 v_{x}= -kt/m+ c

From the given problem, we have

v_{x}=v_{0 }cosθ_{0} at t=0

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

ln (v_{x}/ v_{0 }cosθ_{0 })= -kt/m

or v_{x}=v_{0}cosθ_{0}e^{-kt/m}

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

Hence,

m dv_{y}/dt= -kv_{y}-mg

or, dv_{y}/(kv_{y}+mg)= -k/m dt

Integrating both sides,

ln|kv_{y}+mg|=-kt/m+c

At t=0, v_{y}=v_{0}sinθ_{0}

Hence,

kv_{y}+mg=(kv_{0}sinθ_{0}+mg) e^{-kt}

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