Yeah - I kinda get that it is a pole-vaulting exercize. "Salto Fosbury" translates into English as "Fosbury Flop".
In the model - a pole is set at an angle to the horizontal and travels left-to-right at some constant initial speed u.
When the bottom of the pole is in contact with the bottom of the wall, the horizontal motion stops and the pole rotates about the bottom end.
You've seen that this is a conservation of angular momentum problem - and you seem to have a handle on it.
The pole has length ##l##, mass ##m##, and initial velocity ##\vec v_i=u\hat\imath## (i.e. speed ##u## in the ##+x## direction.)
It has initial linear momentum was ##\vec p_i=mu\hat\imath## ... you follow so far?
Your question appears to be "what happens to this?"
Particularly, what happens to the part that does not turn into angular momentum?
Consider:
The part that turns into angular momentum is the component of the initial momentum that is perpendicular to the pole. ##p_\perp = mu\sin\theta##
... that's the magnitude - but momentum is a vector.
So what is the momentum that is left over?
Which direction does it point in? (A sketch will be enough to see.)
Consider also: Is the pole acted on by a force during the vault?