A projectile of mass 200 g is launched in a viscous medium at an angle $60^{\circ}$ with the horizontal, with an initial velocity of $270 \mathrm{~m} / \mathrm{s}$. It experiences a viscous drag force $\vec{F}=-c \vec{v}$ where the drag coefficient $c=0.1 \mathrm{~kg} / \mathrm{s}$ 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

Solution

$\overrightarrow{\mathrm{F}}_{\mathrm{net}}=\mathrm{m} \frac{\mathrm{~d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}$ $\mathrm{mg}+\overrightarrow{\mathrm{F}}=\frac{\mathrm{md} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}$ $\mathrm{mg}-\mathrm{C} \overrightarrow{\mathrm{v}}=\frac{\mathrm{md} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}$ Horizontal direction $-\mathrm{Cv}_{\mathrm{x}}=\frac{\mathrm{mdv}_{\mathrm{x}}}{\mathrm{dt}}$ $\begin{aligned} & -\frac{C}{m} \int_0^t d t=\int_{v_{0 x}}^{v_x} \frac{d v_x}{v_x} \\ & -\frac{t}{2}=\ell n \frac{v_x}{v_{0 x}} \\ & \frac{d x}{d t}=v_x=v_{0 x} e^{-t / 2} \\ & \int_0^{S_x} d x=v_{0 x} \int_0^t e^{-t / 2} d t \\ & S_x=2 v_{0 x}\left(1-e^{-t / 2}\right) \\ & \text { at } t=2 \sec \end{aligned}$ $\begin{aligned} & \mathrm{S}_{\mathrm{x}}=2 \times 270 \times \cos 60^{\circ}\left[1-\frac{1}{\mathrm{e}}\right] \\ & \mathrm{S}_{\mathrm{x}}=270\left(1-\frac{1}{2.7}\right) \\ & =\frac{270}{2.7} \times(1.7) \\ & =170 \mathrm{~m} \\ & \mathrm{~S}_{\mathrm{x}}=170 \mathrm{~m}\end{aligned}$

Hence, the correct answer is 170.