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

Options

A. `(Hu)/((H-h))`
B. `(Hu)/((H+h))`
C. `((H-h))/(Hu)`
D. `((H+h))/(Hu)`

Answer

A

Explanation

`V=(dy)/(dt)=` velocity of the edge of the shadow on the ground
`mu=(dx)/(dt)=` speed of man.
`(H)/(y)=(h)/(y-x)`
`Hy-Hx=hy`
`y(H-h)=Hx`
differentiating 60th side
`(dy)/(dt)(H-h) =H(dx)/(dt)`
`rArr V=(Hmu)/(H-h)`