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 4cm/s, is
A) $4\sqrt{3}$m
B) $5\sqrt{3}$m
C) 6m
D) 8m
Explanation
(c)
Let $AB = x$m, $BC = y$m and $AC = 10$m
$\therefore\ \ \ x^{2} + y^{2} = 100$ …(i)
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On differentiating w.r.t $t$,m we get
$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$
$\Rightarrow \ \ 2x(3) - 2y(4) = 0$
$\Rightarrow \ \ \ x = \frac{4y}{3}$
On putting this value in Eq. (i), we get
$\frac{16}{9}y^{2} + y^{2} = 100$
$\Rightarrow \ \ \ \ y^{2} = \frac{100 \times 9}{25} = 36\ \ \ \Rightarrow \ \ \ y = 6$m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.