A projectile of mass 200 g is launched in a viscous medium at an angle with the horizontal, with an initial velocity of 270 m/s. It experiences a viscous drag force where the drag coefficient c = 0.1 kg/s and is the instantaneous velocity of the projec

Question

A projectile of mass 200 g is launched in a viscous medium at an angle $60 °$ with the horizontal, with an initial velocity of 270 m/s. It experiences a viscous drag force $\vec{F} = - c \vec{v}$ where the drag coefficient c = 0.1 kg/s and $\vec{v}$ is the instantaneous velocity of the projectile. The projectile hits a vertical wall after 2 s. Taking e = 2.7, the horizontal distance of the wall from the point of projection (in m) is ______. [2025]

Smart Achievers · Accepted Answer

(170)

$Net force$

${\vec{F}}_{net} = m \frac{d \vec{v}}{d t}$

$m \vec{g} + \vec{F} = m \frac{d \vec{v}}{d t}$

$m \vec{g} - c \vec{v} = m \frac{d \vec{v}}{d t}$ $[∵ {\vec{F}}_{drag} = - c \vec{v} given]$

$Horizontal direction$

$- c v_{x} = m \frac{d v_{x}}{d t}$

$- \frac{c}{m} \int_{0}^{t} d t = \int_{v_{0 x}}^{v_{x}} \frac{d v_{x}}{v_{x}} \Rightarrow - \frac{t}{2} = \ln \frac{v_{x}}{v_{0 x}}$

$\frac{d x}{d t} = v_{x} = v_{0 x} e^{- t / 2}$

$\int_{0}^{S_{x}} d x = v_{0 x} \int_{0}^{t} e^{- t / 2} d t$

$S_{x} = 2 v_{0 x} (1 - e^{- t / 2})$

$At t = 2 s$

$S_{x} = 2 \times 270 \times \cos 60 ° [1 - \frac{1}{e}]$

$S_{x} = 270 (1 - \frac{1}{2.7}) = \frac{270}{2.7} \times 1.7 = 170 m$

Solution

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