$\text{A projectile of mass}200g\text{is launched in a viscous medium at an angle}\mathrm{ 60}°\text{with the horizontal, with an initial}$

$\text{velocity of}\mathrm{ 270}\frac{m}{s}.\text{It experiences a viscous drag force}\stackrel{ \to }{F}=-c\overrightarrow{v}\text{where the drag coefficient}c=0.1\frac{\mathrm{kg}}{s}\text{ and }\overrightarrow{v}\text{ is the}$

$\text{instantaneous velocity of the projectile.}\text{The projectile hits a vertical wall after}\mathrm{ 2}s.\text{Taking }e=2.7,\text{the}$

$\text{horizontal distance of the wall from the point of projection (in m) is}\text{\_\_\_\_\_}$

The correct answer is 170.

Step-by-step Explanation:

Let us analyze the horizontal motion of the projectile under the action of the viscous drag force.

1. Identify the given values:
Mass of the projectile, $m=200\text{g}=0.2\text{kg}$
Initial speed, $u=270\text{m/s}$
Angle of projection with the horizontal, $\theta =60°$
Viscous drag coefficient, $c=0.1\text{kg/s}$
Time of travel, $t=2\text{s}$
Base of natural logarithm, $e=2.7$

2. Find the initial horizontal component of velocity:
The horizontal component of the initial velocity ($u_x$) is given by:
$u_x=u\cos \theta$
Substituting the given values:
$u_x=270\times \cos (60°)=270\times 0.5=135\text{m/s}$

3. Derive the expression for horizontal velocity as a function of time:
The drag force is opposing the motion and is given by $\overrightarrow{F}=-c\overrightarrow{v}$. Therefore, the equation of motion along the horizontal x-axis (where gravity has no component) is:
$m\frac{d{v}_x}{dt}=-c{v}_x$
Rearranging the variables to integrate:
$\frac{d{v}_x}{v_x}=-\frac{c}{m}dt$
Integrating from $t=0$ (where $v_x=u_x$) to $t$ (where $v_x=v_x\left(t\right)$):
$\ln \left(\frac{v_x\left(t\right)}{u_x}\right)=-\frac{c}{m}t$
Taking the exponential of both sides:
$v_x\left(t\right)=u_x{e}^{-\frac{c}{m}t}$

4. Calculate the horizontal displacement:
The horizontal distance $x$ covered by the projectile in time $t$ is the integral of the horizontal velocity:
$x={\int }_0^t{v}_x\left(t\right)dt={\int }_0^t{u}_x{e}^{-\frac{c}{m}t}dt$
Integrating this yields:
$x=\frac{u_{x}m}{c}$

5. Substitute numerical values to find the distance:
Let us evaluate the exponential term first:
$\frac{c}{m}t=\frac{0.1}{0.2}\times 2=0.5\times 2=1.0$
Thus, the term $e^{-\frac{c}{m}t}$ simplifies to $e^{-1}=\frac{1}{e}=\frac{1}{2.7}$.
Now, substitute all values into the displacement equation:
$x=\frac{135\times 0.2}{0.1}\left(1-\frac{1}{2.7}\right)$
$x=\frac{27}{0.1}\left(\frac{2.7-1}{2.7}\right)$
$x=270\times \left(\frac{1.7}{2.7}\right)$
$x=10\times 17=170\text{m}$

Thus, the horizontal distance of the wall from the point of projection is 170 meters.