Another collision

as it retraces its path clearly the ball will fall normally on the plane :P

let the angle with which it was thrown w.r.t the plane be β

we have .. vcosβ-gsinθt=0 thus t=vcosβ/gsinθ

further .. 0=vsinβt-gcosθt²/2 thus t=2vsinβ/gcosθ

further as distance =L

vcosβt-gsinθt²/2=l

solving these equations we get v = √(2lg*(cos²Î¸/4 + sin²Î¸)/sinθ