Question

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

• A. $\frac{hv}{(H-h)}$
• B. $\frac{Hv}{(H-h)}$
• C. $\frac{hv}{(H+h)}$
• D. $\frac{(H-h)}{hv}$

Answer 1

The correct option is B $\frac{Hv}{(H-h)}$

Let the situation be as shown below

Here, $OL$ is the lamp post, $MP$ is the man who is walking towards post with the velocity $v$.
$x$ is the length of the shadow and $y$ is the position or distance of man from the post.
So from geometry, $\Delta LOS\text{and}\Delta MPS$ are similar
$\Rightarrow \frac{H}{x}=\frac{h}{(x-y)}\Rightarrow (H-h)x=Hy$
Differentiating L.H.S and R.H.S w.r.t time, we get
$\Rightarrow (H-h)\frac{dx}{dt}=H\frac{dy}{dt}$
So, let velocity of edge of the shadow be $v_s$ which is $\frac{dx}{dt}$
and, $\frac{dy}{dt}$ is the velocity of man i.e. $v$
$\therefore$ we have
$(H-h)v_s=Hv\Rightarrow v_s=\frac{Hv}{(H-h)}$