physics
posted by opy .
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.

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 xL
similar triangles:
x/6=(xL)/1.8
or 1.8x6x=6L
4.2 dx/dt=6dL/dt
given: dL/dt=7, solve for dx/dt 
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/(yx),
6/1.8 = y/(yx),
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.
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