Answer:
0.675 m/s
Step-by-step explanation:
Let height of shadow= y,CD=x
Height of man=2 m
Speed of man= [tex]\frac{dx}{dt}=1. 8 m/s[/tex]
[tex]\triangle ABD\sim\triangle ECD[/tex]
Therefore, [tex]\frac{AB}{EC}=\frac{BD}{CD}[/tex]
[tex]\frac{y}{2}=\frac{12}{x}[/tex]
[tex]xy=24[/tex]
Differentiate w.r.t t
[tex]x\frac{dy}{dt}+y\frac{dx}{dt}=0[/tex]
[tex]x\frac{dy}{dt}=-y\frac{dx}{dt}[/tex]
[tex]\frac{dy}{dt}=-\frac{y}{x}\frac{dx}{dt}[/tex]
When the man is 4 m from the building
Then, we have x=12-4=8 m
[tex]\frac{dx}{dt}=1.8 m/s[/tex]
Substitute the values in above equation then, we get
[tex]8y=24[/tex]
[tex]y=\frac{24}{8}=3[/tex]
Substitute the values then we get
[tex]\frac{dy}{dt}=-\frac{3}{8}\times 1.8=-0.675 m/s[/tex]
Hence, the length of his shadow on the building decreasing at the rate 0.675 m/s.
