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 ______.

Mass of the projectile: $$m = 200 \text{ g} = 0.2 \text{ kg}$$

The drag force is linear: $$\vec F = -c\vec v,$$ with $$c = 0.1 \text{ kg s}^{-1}.$$
Hence for any component of velocity, $$m\dfrac{dv}{dt} = -cv \; \Rightarrow \; \dfrac{dv}{dt} = -\dfrac{c}{m}\,v.$$

Horizontal component
Initial horizontal speed: $$v_{x0} = u\cos 60^{\circ} = 270 \times 0.5 = 135 \text{ m s}^{-1}.$$

With $$\dfrac{c}{m} = \dfrac{0.1}{0.2} = 0.5 \text{ s}^{-1},$$ the differential equation becomes
$$\dfrac{dv_x}{dt} = -0.5\,v_x.$$

Solution for velocity:
$$v_x(t) = v_{x0}\,e^{-0.5t} = 135\,e^{-0.5t}.$$

Horizontal displacement up to time $$t$$:
$$x(t) = \int_0^{t} v_x(t')\,dt' = 135\int_0^{t} e^{-0.5t'}dt' = 135\left[\dfrac{-1}{0.5}e^{-0.5t'}\right]_0^{t}.$$
This simplifies to
$$x(t) = \dfrac{135}{0.5}\bigl(1-e^{-0.5t}\bigr) = 270\bigl(1-e^{-0.5t}\bigr).$$

The projectile hits the wall at $$t = 2 \text{ s}$$, so
$$x(2) = 270\bigl(1-e^{-1}\bigr).$$

Using the given value $$e = 2.7$$,
$$e^{-1} = \dfrac{1}{2.7} \approx 0.37037,$$
$$1 - e^{-1} \approx 1 - 0.37037 = 0.62963.$$

Therefore
$$x(2) = 270 \times 0.62963 \approx 170 \text{ m}.$$

Hence the vertical wall is about $$170 \text{ m}$$ from the point of projection, which lies within the accepted range 167 - 171 m.