Question

A man of height 180 centimetre is moving away from a lamp post at a rate of 1.2 m/s. If the height of the lamp is 4.5m , find the rate at which
(I) It's shadow is lengthening
(II) Tip of the shadow is moving

Detailed explanation needed !!!
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Answer 1

Answer:

2 m/sec

Explanation:

AB = lamp-post,

AB = 4.5 cm

CD = Man.

CD = 180 cm =1.8 m

DE = Shadow

and distance of the tip E of the shadow from the pole = BE

Let BD = X and DE = Y

then BE = X + Y

Now D ABE ~ D CDE ... (By AA test for similar triangles)

AB/CD = BE/DE

45/18 =X + Y / Y

5/2 = X + Y / Y

5Y = 2X + 2Y

3Y = 2X

Y= 2X/3 ===>Y= (2/3) × X

Differentiating w. r. to ’ t ’.

dy/dt = 2/3 × dx/dt (given)

dx/dt = 1.2 m/sec

therefore, dy/dt = (2/3) × 1.2

=0.8 m/sec

 The shadow is lengthening at the rate 0.8 m / sec.

Also BE = X + Y

 Differentiating w. r. to ’ t’

d(BE)/dt = (dx/dt) + (dy/dt)

d(BE)/dt = 1.2 + 0.8

= 2 m/sec

 The rate at which the tip of the shadow is moving away from the lamp-post is 2 m/sec.

Answer 2

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A man of height 180 cm is moving away from the lamp post at the rate of 1.2m per second. If the height of the lamp post is 4.5m.