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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

Aditya
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2 Answers2

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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.

Quanto
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  • 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? – Aditya Aug 21 '19 at 13:16
  • 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. – Quanto Aug 21 '19 at 13:31
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I have got a bit different answer, can someone point my error please? my answer is Hv/(H-h) instead of hv/(H-h)

Thanks!

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

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

Aditya
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Aditya Garg
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