Question 263: Laws of Motion

Question

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 ___________.

Accepted Answer

$$ \vec{F}_{n e t}=m \frac{d \vec{v}}{d t} $$
$$ \mathrm{m} \overrightarrow{\mathrm{~g}}+\overrightarrow{\mathrm{F}}=\frac{\mathrm{md} \overrightarrow{\mathrm{v}}}{\mathrm{dt}} $$
$$ \mathrm{m} \overrightarrow{\mathrm{~g}}-\mathrm{C} \overrightarrow{\mathrm{v}}=\frac{\mathrm{md} \overrightarrow{\mathrm{v}}}{\mathrm{dt}} $$
Horizontal direction
$$ -\mathrm{Cv}_{\mathrm{x}}=\frac{\mathrm{mdv}_{\mathrm{x}}}{\mathrm{dt}} $$
$\begin{aligned} & -\frac{C}{m} \int_0^t d t=\int_{v_{0 x}}^{v_x} \frac{d v_x}{v_x} \ & -\frac{t}{2}=\ln \frac{v_x}{v_{0 x}} \ & \frac{d x}{d t}=v_x=v_{0 x} e^{-t / 2} \ & \int_0^{s_x} d x=v_{0 x} \int_0^t e^{-t / 2} d t \ & S_x=2 v_{0 x}\left(1-e^{-t / 2}ight)\end{aligned}$
At t = 2 sec
$\begin{aligned} & S_x=2 	imes 270 	imes \cos 60^{\circ}\left[1-\frac{1}{\mathrm{e}}ight] \\ & S_x=270\left(1-\frac{1}{2.7}ight) \\ & =\frac{270}{2.7} 	imes(1.7) \\ & =170 \mathrm{~m} \\ & S_x=170 \mathrm{~m}\end{aligned}$

Laws of Motion

Integer / Numeric Answer Type