Question

# A $$2m$$ ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate $$25$$ cm/sec, then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is $$1$$m above the ground is?

A
253
B
25
C
253
D
253

Solution

## The correct option is C $$\dfrac{25}{\sqrt{3}}$$$$x^2+y^2=4\left(\dfrac{dy}{dt}=-25\right)$$$$x=\dfrac{dx}{dy}+y\dfrac{dy}{dt}=0$$$$\sqrt{3}\dfrac{dx}{dt}-1(25)=0$$$$\dfrac{dx}{dt}=\dfrac{25}{\sqrt{3}}$$cm/sec.Mathematics

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