A 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

h=1.8 m, H=6 m, v=7 m/s u=?

x is the distance of man to lamp post,
y is the distance from the tip of the shadow to the lamp post.

From the similar triangles
H/h = y/(y-x),
6/1.8 = y/(y-x),
6y -6x=1.8y,
4.2 y=6 x,
0.7y=x,
d{0.7y}/dt=dx/dt,
0.7(dy/dt)=dx/dt,
0.7u =v,
u=v/0.7=7/0.7=10 m/s.