Question

A ladder $5m$ long is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall, at the rate of $2m/sec$. The speed at which its height on the wall decreases when the foot of the ladder is $4m$ away from the wall is

• A. $\frac{3}{8}m/sec$
• B. $\frac{8}{3}m/sec$
• C. $\frac{5}{3}m/sec$
• D. $\frac{2}{3}m/sec$

Answer 1

Answer: (B)

Let $AB$ be the ladder of length 5 m.
We are given, $\frac{dx}{dt}=2m/sec$
In $\Delta ABC,$
$A{B}^2=A{C}^2+B{C}^2$
$\Rightarrow (5{)}^2=x^2+h^2$ ....(i)
Differentiating (i) w.r.t. t,
we get
$0=2x\frac{dx}{dt}+2h\frac{dh}{dt}$
$\Rightarrow \frac{dh}{dt}=\frac{-x}{h}\frac{dx}{dt}$ .....(ii)
Now, from (i), when $x=4$
$h^2=25-16\Rightarrow h^2=9\Rightarrow h=3m$
From (ii)
$\left[\frac{dh}{dt}\right]=\frac{-4}{3}\times 2=\frac{-8}{3}$
$\because$ The negative sign shows the height decreases and decreasing rate is $\frac{8}{3}m/sec$