The velocity of the image of a moving object in situation shown in figure is (given velocity w.r.t ground)

The velocity of the image of a moving object in situation shown in fig

1.	The velocity of the image of a moving object in situation shown in figure is (given velocity w.r.t ground)
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :We know for the perpendicular component <br/> `(vecV_(1//M))_(bot)=-(vecV_(o//M))_(bot)` <br/> `(V_(IG))_(bot)-(V_(<a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>))_(bot)=-[(V_(<a href="https://interviewquestions.tuteehub.com/tag/og-583305" style="font-weight:bold;" target="_blank" title="Click to know more about OG">OG</a>))_(bot)-(V_(MG))_(bot)]` <br/> `(V_(MG))_(bot)= ((V_(OG))_(bot)+(V_(IG))_(bot))/2 <a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> -30=(10+(V_(IG))_(bot))/2 implies` <br/> `(V_(IG))_(bot)=-70` <br/> THe parallel component of velocity of image is same as that of <a href="https://interviewquestions.tuteehub.com/tag/object-11416" style="font-weight:bold;" target="_blank" title="Click to know more about OBJECT">OBJECT</a> w.r.t ground . It does not depend on the velocity of mirror. So `(V_(IG))_(II)=5m//s`</body></html>