Answer
$5.2\space m/s$, $52^{\circ}$ west of south
Work Step by Step
Let's take,
The velocity of the puck relative to Mario = $V_{PM}$
The velocity of the puck relative to ice = $V_{PI}$
The velocity of the ice relative to Mario = $V_{IM}$
We can write,
$V_{PM}=V_{PI}+V_{IM}=V_{PI}-V_{MI}$
Please see the attached image.
By using the Pythagorean theorem, we can get.
$V_{PM}=\sqrt {V_{(PM)x}^{2}+V_{(PM)y}^{2}} =\sqrt {(V_{PI}-V_{MI})_{x}^{2}+(V_{PI}-V_{MI})_{y}^{2}}$
$V_{PM}=\sqrt {(-11sin22^{\circ}+0)^{2}+(-11cos22^{\circ}+7)^{2}}\space m/s=5.2\space m/s$
By using trigonometry, we can get.
$tan\theta=\frac{4.1\space m/s}{3.2\space m/s}=>\theta=tan^{-1}(1.28)=52^{\circ}$ west of south
