It is given that the length of the ladder is: $AB=5 \mathrm{m}$

From the geometry,

$x^2+y^2=5^2$

Differentiate the above equation.

$2x\frac{dy}{dx}+2y\frac{dy}{dx}=0$

$x\frac{dy}{dx}+y\frac{dy}{dx}=0$

It is given that,

$\frac{dx}{dt}=3,\frac{dy}{dt}=-4$

Substitute the values in the equation $\left(\mathrm{I}\right)$.

$3 x-4 y=0$

$3\sqrt{25-y^2}-4y=0$

$3\sqrt{25-y^2}=4y$

Squaring both sides,

$9\left(25-y^2\right)=16y^2$

By solving,

$y=3$