The answer are:
#v_(0x)=17.2m/s#,
#v_(0y)=24.5m/s#,
final velocity #v=30m/s#,
final angle #theta=125°#.
The motion of a projectile is a composition of two indipendent motions. The first, horizontal, is an uniform motion (#vecv# is costant), and the second, vertical, is an accelerated motion (with #a=-g#, #g# is the gravity acceleration).
So we can consider, for now, the vertical motion. The low is:
#v=v_0+at#
that becomes, in this case,
#v=v_0-g t#.
The projectile goes up until his velocity is #0#, and the time of fly is #2.5s#, so:
#v_0=v+g t=0+9.8m/s^2*2.5s=24.5m/s#,
that is the vertical component we are searching.
#v_(0y)=24.5m/s#.
Given #theta# is the angle of #vecv_0# with respect to the horizontal, we know that:
#tantheta=v_(0y)/v_(0x)rArrv_(0x)=v_(0y)/tanthetarArrv_(0x)=(24.5m/s)/tan(55°)~=17.2m/s#.
So:
#v_0=sqrt(24.5^2+17.2^2)~=30m/s#.
When the projectile lands, his velocity is #v=85m/s#, and the angle with respet of the horizontal is #theta=180°-55°=125°#, because the motion is specular.
