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 $F = -c v$ where the drag coefficient $c = 0.1$ kg/s and 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 ______.

Step 1: Identify the given parameters.

Convert the given values to SI units.

$$m = 0.2\,\text{kg}$$ $$v_0 = 270\,\text{m/s}$$ $$\theta = 60^\circ$$ $$c = 0.1\,\text{kg/s}$$ $$t = 2\,\text{s}$$ $$e = 2.7$$

Step 2: Find the initial horizontal velocity component.

Calculate the horizontal component of the initial velocity.

$$v_{0x} = v_0 \cos 60^\circ$$ $$v_{0x} = 135\,\text{m/s}$$

Step 3: Set up the equation of motion in the horizontal direction.

The horizontal motion is affected only by the viscous drag force.

$$m \frac{dv_x}{dt} = -c v_x$$

Step 4: Solve for the horizontal velocity.

Integrate the differential equation.

$$v_x(t) = v_{0x} e^{-\frac{c}{m}t}$$

Step 5: Find the horizontal displacement.

Integrate the velocity function with respect to time.

$$x(t) = \frac{m v_{0x}}{c} \left(1 - e^{-\frac{c}{m}t}\right)$$

Step 6: Calculate the ratio of mass to drag coefficient.

Evaluate the time constant parameter.

$$\frac{m}{c} = 2\,\text{s}$$

Step 7: Calculate the horizontal distance at $t = 2\,\text{s}$.

Substitute the values into the displacement equation.

$$x(2) = 2 \times 135 \left(1 - e^{-1}\right)$$ $$x(2) = 270 \left(1 - \frac{1}{2.7}\right)$$ $$x(2) = 270 \times \frac{1.7}{2.7}$$ $$x(2) = 170\,\text{m}$$

Answer: 170