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|dw:1444353532318:dw|

you are told \(x'=3\)

when \(t=2\) how far is the ladder away form the wall?

Is there a formula to use on how to calculate that answer?

|dw:1444353987783:dw|

|dw:1444354010477:dw|

not really a formula...

So x would equal 11 at t=2
and that means that h would be about 11.6.

|dw:1444354236271:dw|

put those in last
use pythagoras to find the relation between \(x,h\) and \(16\)

I ended up getting about 2.84

you're welcome

Obviously in the real world, the speed would hit some limit just before the ladder hits the ground

Thank you so much!