22m ladder leans against wall. Bottom of ladder is 3m from wall at t=0, and starts to slide away from wall at 1m/sec.

Question

Find the velocity of the top of the ladder at t=2 seconds.

Answer 1

after 2 seconds the bottom of the ladder is 5 meters away from the wall, so the top of the ladder is sqr(22^2-5^2)= sqr(459) = about 21.4 high

this is a related rates problem

it's a right triangle

l^2 = h^2 + b^2 take the derivative

2ll' = 2hh' + 2bb' divide by 2

ll' = hh' + bb' l'= 0

0 = hh' + bb'

0 = 21.4 h' + 5(2)

h' = -10/21.4= -5/10.7 meters per second

or more exactly h' =-10/sqr459