Relative to another coordinate system mathrm-S- -denoted by single prime- moving with a vertical velocity vec-mathrm-v-0- the equation of motion of the object becomesdisplaystyle mathrm-m-frac-mathrm-d-vec-mathrm-v-mathrm-d-mathrm-t-mathrm-m-vec-mathrm-g-mathrm-b-vec-mathrm-v-mleft - frac-dvec-mathrm-v-dt- right -mvec-g-b-vec-mathrm-v-vec-mathrm-v-0-mleft - frac-dvec-mathrm-v-dt- right -m-vec-g-vec-mathrm-v-0-bvec-mathrm-v-mleft - left - frac-dvec-mathrm-v-dt-vec-mathrm-v-0- right - right -mvec-g-b-vec-mathrm-v-vec-mathrm-v-0-

Question

Toppr Admin · Accepted Answer

The acceleration due to gravity will be the same in both systems
$\to v +_{0} =^{'}$
$\Rightarrow \to v =^{'} -_{0}$
$\frac{d \to v}{d t} = \frac{d \to v}{d t} - \frac{d_{0}}{d t}$
$\frac{d \to v}{d t} = \frac{d^{'}}{d t}$
$\Rightarrow m \frac{d \to v}{d t} = m (\frac{d^{'}}{d t})$
$\Rightarrow m (\frac{d^{'}}{d t}) = - m g - b (\to v^{'} -_{0})$

Relative to another coordinate system S′ (denoted by single prime) moving with a vertical velocity →v0, the equation of motion of the object becomesm(d→vdt)=−m→g−b→vm(d→vdt)=−m→g−b(→v−→v0)m(d→vdt)=m(→g−→v0)−b→vm[(d→vdt−→v0)]=−m→g−b(→v+→v0)

The acceleration due to gravity will be the same in both systems→v+→v0=→v′⇒→v=→v′−→v0d→vdt=d→vdt−d→v0dtd→vdt=d→v′dt⇒md→vdt=m(d→v′dt)⇒m(d→v′dt)=−mg−b(→v′−→v0)