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I guess u meant coefficient of restitution, by coefficient of friction!!
First we should know that coefficient of restitution, e= relative speed of seperation  = RSS/RSA
          relative speed of approach
(Wish i could help you with the visuals but i don''t know how to)
Never mind, lets continue this way..
before collision
let the ball fall at angle of 37degree, with the normal, then dividing the velocity into components we have-
vsin37 along verticall downward, direction is -j^, and
vcos37  along horizontal towards the wall, i.e direction along -i^
 
after collision
let the velocity be u, alog vertically downward it is     usin37 -j^
and along horizontal it is ucos37 +i^
 
Now vertical velocity is not affect during collision, because only horizontal velocity takes part in collision.
therefore, usin37=vsin37
or, u=v
therefore horizontal velocity after collision is,      vcos37
applying very first formulae, we have e= RSS/RSA 
or,  e=vcos37/vco37=1
therefore, we can say collision is perfectly elastic.
 
 
Now we have a shortcut also, when we have angles of collision and refelection, that might help (:D)-
e=tanΘ1/tanΘ2
here tanΘ1 is intial angle and tanΘ2 isfinal angle..
hope this helps :D
AJ
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