Question

man of height 1.8m walks away from a lamp at a height of 6m.If the man's speed is 7m/s,find the speed in m/s at which the tip of the shadow moves​

Answer 1

draw a right triangle: vertical 6m, horizontal leg x

Now within that triangle, another vertical 1.8 m, and the horizontal distance to the lamppost is L, so the length to the tip of shadow is x-L

similar triangles:

x/6=(x-L)/1.8

or 1.8x-6x=-6L

4.2 dx/dt=6dL/dt

given: dL/dt=7, solve for dx/dt

Answer 2

Answer:

Heya

Explanation:

The shadow triangle and the men’s triangle with lamp would be simlar.

Hence let x be the distance of tip of the shadow from man and y be the distance of man from the lamp at any instant.

Then,

from similar triangles we get,

1.8/x = 6(x+y)

or

7x = 3y........................................................(1)

Now from distance – speed formulae:

for man,

7 = y/t => t = y/7 seconds.

for shadow,

v = (x+y)/t => t = (x+y)/v

now time would be same for both therfore,

v = 7x/y + 7

from eqn. (1) we get,

v = 3 + 7 => 10 m/s.