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The base of a 20-foot ladder is being pulled away from a wall at a rate of 12 feet per second. How fast is the top of the ladder sliding down the wall at the instant when the base of the ladder is 12 feet from the wall?

Respuesta :

Answer:

v= 20 ft/s

Explanation:

Given that

x= 20 ft

u= 12 ft/s

y= 12 ft

v=?

From rigid body concept all point on the rod will rotate a about point I.This I point is known as I -center.

Lets take angular speed of the rod is ω  and which is rotating about I.

So

u  =  ω y

v= ω x

From above equation

v/u = x/y

By putting the values

v/12 = 20 /12

v= 20 ft/s

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