Answer
0.632 m/s
Work Step by Step
Let x be the distance from the bottom of the ladder to the wall and h be the height of its top.
The goal is to find $\frac{dh}{dt}$ at $ t=2s $.
We are given that $\frac{dx}{dt}=0.8\, m/s $ and $ x(0)=1.5\,m $
Using Pythagoras' theorem, we have
$ x^{2}+h^{2}=(5\,m)^{2}$
Differentiating both sides with respect to $ t $, we have
$2x\frac{dx}{dt}+2h\frac{dh}{dt}=0$
$\implies \frac{dh}{dt}=-\frac{x}{h}\frac{dx}{dt}=-\frac{x}{h}\times0.8\,m/s $
$ x(2)=x(0)+dt\frac{dx}{dt}=1.5\,m+ 2\,s(0.8\,m/s)=3.1\,m $
$ h(2)=\sqrt {(5\,m)^{2}-(3.1\,m)^{2}}\approx 3.923\,m $
$\frac{dh}{dt}|_{t=2}=-\frac{3.1\,m}{3.923\,m}\times0.8\,m/s=0.632\,m/s $
