Expert: Abe Mantell - 8/14/2010

Question
Hi there,
My question is,
Bob has a ladder that is 5m long resting agaist the top of a 4m high wall. Bob is 3m up the ladder and all of a sudden the ladder begins to slide down the wall. Wendy his wife, measures the top of the ladder as it slides down the wall; it is a constant speed of 2m/s down.
How do i find
a) the time it takes for the ladder to land on the qround
and
b)would the foot of the ladder be traveling at a constant speed?

Answer
Hello Brigette,

My first thought was, "Why is Bob's wife measuring instead of helping
to stop Bob from getting hurt?!!"  ;-)

Anyhow...

a) Since the height of the ladder on the wall was 4 m, and it is
.  falling at 2 m/sec, it will take (4 m)/(2 m/sec) = 2 sec to hit
.  the ground.

b) Let the distance the foot of the ladder is from the wall be x,
.  and the height on the wall be y.  Thus, x^2 + y^2 = 5^2, or
.  x^2 + y^2 = 25.  Now differentiate with respect to 't' (time) to
.  get: 2x*dx/dt + 2y*dy/dt = 0.  Since dy/dt = -2 m/sec, we get
.  2x*dx/dt + 2y*(-2) = 0 ==> dx/dt=2y/x  m/sec.
.  Thus, the rate at which the foot of the ladder is moving (dx/dt)
.  is NOT constant, it depends on x & y!

I hope this helps.

Abe