1. Given data

• Mass of projectile: $m=200\text{ g}=0.2\text{ kg}$
• Initial speed: $u=270\text{ m/s}$
• Projection angle: $60^\circ$
• Drag force: $\vec F=-c\vec v$, with $c=0.1\text{ kg/s}$
• Time to hit wall: $t=2\text{ s}$
• Take $e=2.7$

We need the horizontal distance of the wall.

2. Horizontal motion equation

Since drag is proportional to velocity, along the horizontal direction:

$m\frac{dv_x}{dt}=-cv_x$

So,

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

This gives

$v_x(t)=u_x e^{-ct/m}$

where

$u_x=u\cos 60^\circ=270\cdot \frac12=135\text{ m/s}$

Now,

$\frac{c}{m}=\frac{0.1}{0.2}=0.5\ \text{s}^{-1}$

Hence,

$v_x(t)=135e^{-0.5t}$

3. Horizontal displacement

Horizontal position is

$x(t)=\int_0^t v_x\,dt=\int_0^t 135e^{-0.5t}\,dt$

Using

$\int e^{-0.5t}dt=-2e^{-0.5t}$

we get

$x(t)=135\left[\frac{1-e^{-0.5t}}{0.5}\right]$

$x(t)=270\left(1-e^{-0.5t}\right)$

At $t=2\text{ s}$,

$x(2)=270(1-e^{-1})$

Given $e=2.7$, we use

$e^{-1}=\frac{1}{2.7}$

Therefore,

$x(2)=270\left(1-\frac{1}{2.7}\right)$

$x(2)=270\left(\frac{1.7}{2.7}\right)$

$x(2)=100\times 1.7=170\text{ m}$

4. Final answer

The horizontal distance of the wall is

$\boxed{170\text{ m}}$

5. Comparison with stored answer

Stored correct answer: $167\text{ to }171$

Our answer $170$ lies in this range, so it agrees.