The velocity of the edge of shadow on ground will be.

The answer is $$\frac{hu}{H-h}$$

All I was able to do was draw this measly diagram. It’s an understandably difficult to infer question, if not tough as a whole. Help would be appreciated enter image description here


Let $x$ be the distance of the shadow edge to the post and y the distance of the man to the post. From similar triangles, you could establish $x/(x-y)=H/h$. Then, rearrange it to get

$$ x = \frac{H}{H-h} y $$

And, take the time derivatives on both sides,

$$ x’ = \frac{H}{H-h} y’ =\frac{Hu}{H-h}$$

where $y’$ is the velocity of the man, i.e. $y’=u$, and $x’$ is the velocity of the shadow.

  • $\begingroup$ It’s seems to work, but I am still confused about the final answer you obtained. It’s might be right, perhaps I am being dumb, but could you please elaborate your steps?the part where you said$x’=\frac{h}{H-h}y’$, what does it mean in the context of the question? $\endgroup$ – Aditya Aug 21 at 13:16
  • $\begingroup$ The velocity is defined as the derivative of distance with respect to time. Since x is the distance of the shadow to the post, $x’$ is its velocity. $\endgroup$ – Quanto Aug 21 at 13:31

I have got a bit different answer, can someone point my error please? my answer is Hv/(H-h) instead of hv/(H-h)


$$\frac{x}{x-y}=\frac{H}{h}$$ $$x=\frac{Hx}{h}-{Hy}{h}$$ $$\frac{x(h-H)}{h}=-\frac{Hy}{h}$$ $$x=\frac{Hy}{H-h}$$ $$x=\frac{Hv}{H-h}$$

Where x is position of shadow from lamp post, y is position of man from lamp post, y’ is rate of change of position from of the man

  • $\begingroup$ You’re actually right. Will correct. Thanks $\endgroup$ – Quanto Aug 21 at 14:16
  • $\begingroup$ You were very close, however it’s supposed to be h in the numerator. Regardless, next please try to use Math Jax to write your answers. I have done it for you this time, but you will soon get the hang of it! Keep helping. $\endgroup$ – Aditya Aug 21 at 14:34
  • $\begingroup$ Okayt! thank you $\endgroup$ – Aditya Garg Aug 21 at 15:09

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