3
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.