By using the suvat equations:
(a) [tex]v^2=u^2+2as[/tex] for the vertical components, so
[tex]0^2=(5sin57)^2+2*-9.81*s[/tex]
therefore s=0.896m
However, since the ramp is 1.2m above the ground, the skateboarder is 0.896+1.2=2.096m above the ground.
(b) Work out the time when the skateboarder reaches the highest point
v=u+at for the vertical components, so 0=5sin57-9.81*t, therefore t=0.427s
Now using the horizontal components: s=ut, so s=5cos57*0.427=1.164m