A) \[\frac{hu}{(H-h)}\]
B) \[\frac{Hu}{(H+h)}\]
C) \[\frac{(H-h)}{Hu}\]
D) \[\frac{(H+h)}{Hu}\]
E) \[\left( \frac{H+h}{H-h} \right)u\]
Correct Answer: A
Solution :
Let in same time t, distance covered by man and edge of image be respectively\[x\]and y. \[\tan \theta =\frac{H-h}{x}=\frac{H}{x+y}\] \[\Rightarrow \] \[Hx=Hx+Hy-hx+hy\] \[\Rightarrow \] \[\frac{x}{y}=\frac{(H-h)}{h}\] ?. (1) Now \[x=ut\] and \[y=vt\] \[\Rightarrow \] \[\frac{x}{y}=\frac{u}{v}\] Hence, \[\frac{u}{v}=\frac{(H-h)}{h}\] \[\Rightarrow \] \[v=\frac{hu}{(H-h)}\]You need to login to perform this action.
You will be redirected in
3 sec