Horizontal Distance of Projectile in Viscous Medium with Drag

To find the horizontal distance $x$ from the point of projection to the vertical wall, we analyze the motion of the projectile along the horizontal direction ($x$-axis).

The force equation for horizontal motion is given by: $m \frac{dv_x}{dt} = -c v_x$

where:

• Mass of the projectile, $m = 200\text{ g} = 0.2\text{ kg}$
• Viscous drag coefficient, $c = 0.1\text{ kg/s}$
• Initial speed, $v_0 = 270\text{ m/s}$
• Launch angle, $\theta = 60^\circ$

Rearranging the differential equation: $\frac{dv_x}{v_x} = -\frac{c}{m} dt$

Let $\gamma = \frac{c}{m} = \frac{0.1\text{ kg/s}}{0.2\text{ kg}} = 0.5\text{ s}^{-1}$.

Integrating both sides with initial condition $v_x(0) = v_{0x}$: $v_x(t) = v_{0x} e^{-\gamma t}$

The initial horizontal component of velocity is: $v_{0x} = v_0 \cos 60^\circ = 270 \times \frac{1}{2} = 135\text{ m/s}$

The horizontal distance $x(t)$ traveled in time $t$ is obtained by integrating $v_x(t)$: $x(t) = \int_0^t v_x(t') \, dt' = \int_0^t v_{0x} e^{-\gamma t'} \, dt' = \frac{v_{0x}}{\gamma} \left(1 - e^{-\gamma t}\right)$

For $t = 2\text{ s}$, the exponent is: $\gamma t = 0.5 \times 2 = 1$

Using $e = 2.7$ as given in the problem: $e^{-\gamma t} = e^{-1} = \frac{1}{2.7}$

Substituting these values into the expression for $x(2)$: $x(2) = \frac{135}{0.5} \left(1 - \frac{1}{2.7}\right)$ $x(2) = 270 \times \left(1 - \frac{10}{27}\right)$ $x(2) = 270 \times \frac{17}{27} = 170\text{ m}$

Thus, the horizontal distance of the wall from the point of projection is $170\text{ m}$.