Answer:
When
person is at rest with respect to ground, the rain is coming to him at an angle
\[{{30}^{o}}\]with the vertical i.e., along OB. As
the person moves along OA with velocity 10 m/s, the relative velocity of rain
w.r.t. person is along OC as shown in Fig. 2(HT).5. Here
\[\angle BOC={{30}^{o}}\]
\[\therefore \] Velocity of rain w.r.t. ground \[=\overset{\to
}{\mathop{{{\upsilon }_{rg}}}}\,=\overset{\to }{\mathop{OB}}\,\]
Velocity of person w.r.t. ground\[=\overset{\to
}{\mathop{{{\upsilon }_{pg}}}}\,=\overset{\to }{\mathop{OA}}\,\] where OA = 10
m./s = CB
Velocity of rain w.r.t. person \[=\overset{\to
}{\mathop{{{\upsilon }_{rp}}}}\,=\overset{\to }{\mathop{OC}}\,\]
(a)
In
\[\Delta \,OCB\,\], \[OB=CB/\sin {{30}^{o}}=10/\left( 1/2 \right)=20\]
Velocity
of rain w.r.t. ground = 20m/s
(b)
In
\[\Delta OCB\], \[OC=CB/\tan {{30}^{o}}=10/\left( 1/\sqrt{3}
\right)=10\sqrt{3}\]
\[\therefore
\]Velocity of rain w.r.t. person =\[=10\sqrt{3}m/s\]
You need to login to perform this action.
You will be redirected in
3 sec
