(15) A ladder 10 meter long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at the rate of 1.2 meters per second, find how fast the top of the ladder is sliding down the wall when the bottom is 6 meters away from the wall.
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Given (15) A ladder 10 meter long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at the rate of 1.2 meters per second, find how fast the top of the ladder is sliding down the wall when the bottom is 6 meters away from the wall.
- Now we have a ladder of 10 m length and let suppose vertical distance is y and horizontal distance is x.
- So from Pythagoras theorem we have
- So x^2 + y^2 = l^2
- Or y^2 = 10^2 – 6^2
- Or y^2 = 64
- Or y = 8 m
- Now rate of change of x will be
- 2x dx/dt + 2y dy/dt = 2l dl/dt
- Since length of ladder is constant dl/dt = 0
- So 2 (-1.2) + 16 (dy/dt) = 0
- So - 2.4 + 16 dy/dt = 0
- Or – 2.4 = - 16 dy/dt
- Or 16 dy/dt = 2.4
- Or dy/dt = 2.4 / 16
- Or dy/dt = 0.15
- So the top of ladder is sliding 0.15 m / s
Reference link will be
https://brainly.in/question/2346772
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