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?

Answer 1

Answer:

$\dfrac{h}{H-h}u$

Explanation:

Given:

• h = height of the person
• H = height of the lamp post
• u = velocity of the man moving towards the lamp

Assume:

• v = velocity of the edge of the shadow on the ground
• x = distance of the person from the lamp
• y = distance of the edge of the shadow from the man

According to the given question and the above assumptions, we have

Using the similarity criteria between the two triangles, we have

$\dfrac{x+y}{y}=\dfrac{H}{h}\\\Rightarrow \dfrac{x}{y}+1=\dfrac{H}{h}\\\Rightarrow \dfrac{x}{y}=\dfrac{H}{h}-1\\\Rightarrow \dfrac{x}{y}=\dfrac{H-h}{h}\\\Rightarrow y=\dfrac{h}{H-h}x\\\Rightarrow \dfrac{dy}{dt}=\dfrac{d}{dt}\dfrac{h}{H-h}x\\\Rightarrow \dfrac{dy}{dt}=\dfrac{h}{H-h}\dfrac{dx}{dt}$

Since, the rate of change in position of the person from the lamp post will the velocity of the person and the rate of change of endge of the shadow will be the velocity of the edge of the shadow.

$\therefore v=\dfrac{h}{H-h}u$

Hence, the velocity of the edge of the shadow is $\dfrac{h}{H-h}u$.