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The top of ladder 6m 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 ladder is 4m from the wall ,it is sliding away from the wall at the rate of 0.5m/sec. how fast is the top-sliding downwards at instanse? How far is the foot the wall when it and the top are moving at the same rate?
Dear Rohit,
When base is 4m away, angle with horizontal = t = cos^-1(6/4)=cos^-1(3/2). Since length of ladder is constant, velocity of base and velocity of top along the ladder should be same => 0.5 cos(t) = v_top * cos (90°-t)
=> v_top = 0.5 cot (t) = 0.5 * (2/sqrt(5)) = 1/sqrt(5) m/s
For v_top = v_base, cot(t) = 1 from above equation => t = 45°.Please feel free to post as many doubts on our discussion forum as you can.we will get you the answer and detailed solution very quickly.All the best.Regards,
Askiitians ExpertsGokul Joshi
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