A ladder 10 metres long rests with one end against a vertical wall, the other on the floor. The lower end moves away from the wall at the rate of 2 metres/minute. The rate at which the upper end falls when its base is 6 metres away from the wall 

3/2 metres/min

Let at time t the lower end P of ladder PQ be at a distance x metres from the wall and the upper end Q be at height y from the ground. Then,

$x^2+y^2=10^2$

It is given that $\frac{d x}{d t}=2$

We have to find $\frac{d y}{d t}$ when $x=6$

Putting $x=6$ in $x^2+y^2=10^2$, we get $y=8$

Now,

$x^2+y^2=10^2$

$\Rightarrow 2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0$

$\Rightarrow 6 \times 2+8 \times \frac{d y}{d t}=0$         [Putting x = 6, y = 8 and $\frac{d x}{d t}$ = 2]

$\Rightarrow \frac{d y}{d t}=-\frac{3}{2}$ metres/min