Question

A projectile of mass m is fired into a liquid at an angle ${\theta }_0$ with an initial velocity $v_0$ as shown. If the liquid develops a frictional or drag resistance on the projectile which is proportional to its velocity, i.e., F= -kv where k is a positive constant, determine the x and y components of its velocity at any instant. Also find the maximum distance $x_{max}$ that it travels?

• A. $v_x=v_0\cos {\theta }_0{e}^{-kt/m},v_y=\frac{m}{k}[\frac{k}{m}{v}_0\sin {\theta }_0+g^{-\frac{kt}{m}}-g]X_m=\frac{mv\cos \theta }{k}$
• B. $v_x=v_0\cos {\theta }_0{e}^{-2kt/m},v_y=\frac{m}{k}[\frac{k}{m}{v}_0\sin {\theta }_0+g^{-\frac{kt}{m}}-g]X_m=\frac{mv\cos \theta }{2k}$
• C. $v_x=v_0\cos {\theta }_0{e}^{-kt/m},v_y=\frac{m}{k}[\frac{k}{m}{v}_0\sin {\theta }_0+g^{-\frac{2kt}{m}}-g]X_m=\frac{mv\cos \theta }{k}$
• D. $v_x=v_0\cos {\theta }_0{e}^{-kt/m},v_y=\frac{m}{k}[\frac{k}{m}{v}_0\sin {\theta }_0+g^{-\frac{kt}{m}}-g]X_m=\frac{mv\cos \theta }{4k}$

Answer 1

The correct option is A $v_x=v_0\cos {\theta }_0{e}^{-kt/m},v_y=\frac{m}{k}[\frac{k}{m}{v}_0\sin {\theta }_0+g^{-\frac{kt}{m}}-g]X_m=\frac{mv\cos \theta }{k}$

Given : $u_x=V_{o}cos{\theta }_o$ $u_y=V_{o}sin{\theta }_o$

x direction : $F_x=-k{v}_x$ where $F_x=m{a}_x$

$\therefore$ $a_x=\frac{-k}{m}{v}_x$ where $a_x=\frac{d{v}_x}{dt}$

OR ${\int }_{v_{o}cos{\theta }_o}^{v_x}\frac{d{v}_x}{v_x}={\int }_0^t\frac{-k}{m}dt$

$⟹$ $ln\frac{v_x}{v_{o}cos{\theta }_o}=\frac{-kt}{m}$

$⟹$ $v_x=v_{o}cos{\theta }_o{e}^{-kt/m}$

Also ${\int }_0^{x}dx={\int }_0^t\left(v_{o}cos{\theta }_o\right)e^{-kt/m}dt$ (using $v_x=dx/dt$)

$\therefore$ $x=\left(v_{o}cos{\theta }_o\right)\times \frac{e^{-kt/m}}{-k/m}{|}_0^t$ $⟹x=\frac{m{v}_{o}cos{\theta }_o(1-e^{-kt/m})}{k}$

$⟹$ $x_{max}=\frac{m{v}_{o}cos{\theta }_o}{k}$

y direction : $a_y=-(\frac{k}{m}{v}_y+g)$

${\int }_{v_{o}sin{\theta }_o}^{v_y}\frac{d{v}_y}{(\frac{k}{m}{v}_y+g)}={\int }_0^t-dt$

$⟹$ $\frac{ln(\frac{k{v}_y}{m}+g)}{k/m}{|}_{v_{o}sin{\theta }_o}^{v_y}=-t$

OR $ln\frac{\frac{k{v}_y}{m}+g}{\frac{k}{m}\left(v_{o}sin{\theta }_o\right)+g}=\frac{-kt}{m}$

$⟹$ $v_y=\frac{m}{k}[(\frac{k}{m}{v}_{o}sin{\theta }_o+g)e^{-kt/m}-g]$