Up, Up, and ... Oops!

Referring to the figure, at an instant of time, $l^2=a^2+b^2$ .

$b$ is unchanged with time. So, we get

$\Rightarrow 2l \frac{\mathrm d l}{\mathrm d t} = 2a \frac{\mathrm d a}{\mathrm d t}$

But, $\frac{\mathrm d l}{\mathrm d t} = u$ and $\frac{a}{l} = \cos \theta$ .

And if the upward speed of the moving block is taken as $\frac{\mathrm d a}{\mathrm d t} = v$ ,

$v= \frac{l}{a} \frac{\mathrm d l}{\mathrm d t}$

$\Rightarrow v = \frac{u}{\cos \theta}$