Question

A 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 :

• A. $\frac{Hu}{(H-h)}$
• B. $\frac{\left(Hu\right)}{(H+h)}$
• C. $\frac{(H-h)}{Hu}$
• D. $\frac{(H+h)}{Hu}$

Answer 1

The correct option is A $\frac{Hu}{(H-h)}$

From the figure,

$dy/dt=u$

$tan\theta =h/x=H/(y+x)⟹x=hy/(H-h)$

So, $dx/dt=\frac{h}{H-h}dy/dt=hu/(H-h)$

Now the distance of the edge of shadow from the stationary lamp-post is $x+y$

So, the speed of the edge of the shadow on ground would be,

$dx/dt+dy/dt=u+hu/(H-h)=Hu/(H-h)$