question_answer

A man of height h walks in a straight path towards a latnp post of height H with uniform velocity u. Then the velocity of the edge of the shadow on the ground will be

A)  \[\frac{hu}{(H-h)}\]                      

B)  \[\frac{Hu}{(H+h)}\]

C)  \[\frac{(H-h)}{Hu}\]                      

D)  \[\frac{(H+h)}{Hu}\]

Correct Answer: A

Solution :

                 Let in same time t, distance covered by man and edge of image be respectively \[x\] and \[y.\] \[\tan \theta =\frac{H-h}{x}=\frac{H}{x+y}\] \[\Rightarrow \]               \[Hx=Hx+Hy-hx+hy\] \[\Rightarrow \]               \[\frac{x}{y}=\frac{(H-h)}{h}\] Now      \[x=ut\] and        \[y=vt\] \[\Rightarrow \]               \[\frac{x}{y}=\frac{u}{v}\] Hence, \[\frac{u}{v}=\frac{(H-h)}{h}\] \[\Rightarrow \]               \[v=\frac{hu}{(H-h)}\]