A projectile of mass $m$ is fired into a liquid at an angle $θ_{0}$ with an initial velocity $v_{0}$ as shown. If the liquid develops a frictional or drag resistance on the projectile which is proportional to its velocity, i.e. $F = - k v$ where $k$ is a positive constant, determine the $x$ and $y$ components of its velocity at any instant. Also find the maximum distance $x_{\max}$ that it travels?

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$x_{m} = \frac{v_{0} \cos θ_{0}}{k}$

$x_{m} = \frac{m v_{0} \cos θ_{0}}{k}$

$x_{m} = \frac{m \cos θ_{0}}{k}$

$x_{m} = \frac{m v_{0} \cos θ_{0}}{2}$

answer is B.

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Detailed Solution

$a_{x} = \frac{d v_{x}}{d t} = - \frac{k v \cos θ}{m t} = - \frac{k}{m} v_{x} ∴ \frac{d v_{x}}{v_{x}} = - \frac{k}{m} d t or \int_{v_{0}}^{v_{x}} \cos θ_{0} \frac{d v_{x}}{v_{x}} = - \frac{k}{m} \int_{0}^{t} d t v_{x} = v_{0} \cos θ_{0} e^{- \frac{k}{m}}$

A projectile of mass m is fired into a liquid at an angle theta_0 with an initial velocity v_0 as shown. If the liquid develops a frictional or drag resistance on the

Similarly,
$a_{y} = \frac{d v_{y}}{d t} = - \frac{k v \sin θ}{m} - g = - (\frac{k}{m} v_{y} + g) or \int_{v_{0}}^{v_{y}} \sin θ_{0} \frac{d v_{y}}{\frac{k}{m} v_{y} + g} = - \int_{0}^{t} d t or \frac{m}{k} {[\ln (\frac{k}{m} v_{y} + g)]}_{v_{0} \sin θ_{0}}^{v_{y}} = - t$

or $\frac{(\frac{k}{m} v_{y} + g)}{(\frac{k}{m} v_{0} \sin θ_{0} + g)} = e^{- \frac{k}{m} t} - - - - - (i)$
or $v_{y} = \frac{m}{k} [(\frac{k}{m} v_{0} \sin θ_{0} + g) e^{- \frac{k}{m}} - g] - - - - - (i i)$
Eq. (i) can be written as
$\frac{d x}{d t} = v_{0} \cos θ_{0} e^{- \frac{k}{m} t}$
or $\int_{0}^{x} d x = v_{0} \cos θ_{0} \int_{0}^{t} e^{- \frac{k}{m} t} d t$
or
$x = \frac{m v_{0} \cos θ_{0}}{k} [1 - e^{- \frac{k}{m} t}]$
$x_{m} = \frac{m v_{0} \cos θ_{0}}{k}$
at
$t = \infty$