wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The top of a ladder 6 metres long is resting against a vertical wall on a level pavement, when the ladder begins to slide outwards. At the moment when the foot of the ladder is 4 metres from the wall, it is sliding away from the wall at the rate of 0.5 m/sec. How fast is the top-sliding downwards at this instance?
How far is the foot from the wall when it and the top are moving at the same rate?

Open in App
Solution

Let the bottom of the ladder be at a distance of x m from the wall and its top be at a height of y m from the ground.

Here,

x2+y2=362xdxdt=-2ydydt ...1When x=4, y=36-16=252×4×0.5=-2×25dydt dxdt=0.5 m/secdydt=-15m/secFrom eq. (1), we get2xdxdt=-2ydydt dxdt=dydt x=-ySubstituting x=-y in x2+y2=36, we get x2+x2=36x2=18x=32 m

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Linear Dependency of Vectors
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon