The rate is #65/6 "m/min"=10 5/6 "m/min"~~10.83 "m/min"#

This is a nice example of a Related Rates problem. (Perhaps better called by the full name "Related Rates of Change")

**Solution:**

#h# is the distance from the floor to the end of the ladder.

#l# is the length of the ladder.

#(dl)/dt=10"m/min"#

We need to find #(dh)/(dt)#

The relationship between the variables, #h# and #l# is given by the Pythagorean Theorem:

#h^2+5^2=l^2#

We need to find #(dh)/(dt)#

To find the relationship between the rates of change, differentiate (implicitly) with respect to #t#.

#2h(dh)/(dt)+0=2l(dl)/(dt)#

So:

#h(dh)/(dt)=l(dl)/(dt)#

Substitute what we know and solve for the desired value.

We are told that #(dl)/dt=10"m/min"# and we are interested in

the instant when #l=13"m"#, but we also need #h# at that instant.

Using Pythagoras to find #h# when #l=13"m"#

#h^2+5^2=13^2#

#h^2=144#

#h=12"m"#

Now we can finish:

#h(dh)/(dt)=l(dl)/(dt)# so at the instant we were asked about:

#12"m"(dh)/(dt)=13"m"(10"m/min")#

Thus

#(dh)/(dt)=130/12 "m/min"=65/6 "m/min"=10 5/6 "m/min"~~10.83 "m/min"#