Q. A ladder 10 m long rests against a vertical wall with the lower end on the horizontal ground. The lower end of the ladder is pulled along the ground away from the wall at the rate of 3 cm/s. The height of the upper end while it is descending at the rate of 4 cm/s, is
A) $4\sqrt{3}m$
B) $5\sqrt{3}m$
C) $5\sqrt{2}m$
D) 6 m
Explanation
(d)
Let $AB = xm,BC = ym$ and $AC = 10$m
$\therefore\ \ \ \ \ \ \ \ \ \ \ x^{2} + y^{2} = 100$ …(i)
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$\Rightarrow \ \ \ \ \ \ \ \ \ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ 2x(3) - 2y(4) = 0$
$\left\lbrack Given = \frac{dx}{dt} = 3m/s,\frac{dy}{dt} = - 4m/s \right\rbrack$
$\Rightarrow \ \ \ \ \ \ \ \ \ x = \frac{4y}{3}$
On putting this value in Eq.(i),we get
$\frac{16}{9}y^{2} + y^{2} = 100\ \Rightarrow y = 6m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.