Question

A particle of mass $1kg$ is fired with velocity $50m/s$ at an angle of ${60}^o$ from horizontal. It is acted by viscous force of $0.2v$ during its journey. The horizontal distance travelled by it in first $10$ seconds is:-

• A. $90m$
• B. $108m$
• C. $125m$
• D. $213m$

Answer 1

The correct option is B $108m$

$F=0.2va=0.2vm/s^2\frac{dv}{dt}=0.2v{\int }_{25}^v\frac{dv}{v}=-{\int }_0^{10}0.2dtlnv-ln25=-2lnv=ln25-2lnv=1.218v=3.59m/sa=v\frac{dv}{dx}=0.2v{\int }_{25}^{3.59}dv=-{\int }_0^{10}0.2dxx=\frac{25-3.6}{0.2}=107x=108m$