Answer:0.3 m/s
Explanation:
Given
Length of ladder L=10 m
Foot of ladder is Pulled away at the rate of 0.4 m/s
Let the distance of foot of ladder be x m from origin and y be the distance of top of ladder from origin
from Pythagoras we can say that
[tex]x^2+y^2=L^2[/tex]
differentiating w.r.t time we get
[tex]2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0[/tex]
[tex]x\frac{\mathrm{d} x}{\mathrm{d} t}+y\frac{\mathrm{d} y}{\mathrm{d} t}=0[/tex]
[tex]x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}[/tex]
and at x=6m , y=8 m using Pythagoras
[tex]6\times 0.4=-8\times \frac{\mathrm{d} y}{\mathrm{d} t}[/tex]
[tex]\frac{\mathrm{d} y}{\mathrm{d} t}=-0.3 m/s[/tex]
negative indicates that ladder is coming down
