Answer:0.3 m/s
Step-by-step 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
![x^2+y^2=L^2](https://img.qammunity.org/2020/formulas/physics/high-school/2m7u9g57fq8cvx6qghvqarkzhgt88ax9d6.png)
differentiating w.r.t time we get
![2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0](https://img.qammunity.org/2020/formulas/physics/high-school/ktd5paq004gu3like4k0lc00drszszd62w.png)
![x\frac{\mathrm{d} x}{\mathrm{d} t}+y\frac{\mathrm{d} y}{\mathrm{d} t}=0](https://img.qammunity.org/2020/formulas/physics/high-school/u0r7ldw84rlhhku7uqx8dqm4txnznsjw3d.png)
![x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}](https://img.qammunity.org/2020/formulas/physics/high-school/n9wscrek2jsuuyhifcktf2apez0us5dq3z.png)
and at x=6m , y=8 m using Pythagoras
![6* 0.4=-8* \frac{\mathrm{d} y}{\mathrm{d} t}](https://img.qammunity.org/2020/formulas/physics/high-school/mwmb5hl13xx037qzcwztrhdhc5je7u68gq.png)
![\frac{\mathrm{d} y}{\mathrm{d} t}=-0.3 m/s](https://img.qammunity.org/2020/formulas/physics/high-school/ilany8qkhekzvwda0rqxpbfh7u5ke9yt4o.png)
negative indicates that ladder is coming down