Answer
Puck's velocity $=2.2\space m/s,$ angle $=27^{\circ}$ above the +x axis
Work Step by Step
Let's use equations 3.3a,b $V=u+at$ to find the x,y velocity components of the puck.
$\rightarrow V=u+at$
$V_{x}=1\space m/s+2\space m/s^{2}\times 0.5\space s=2\space m/s$
$\uparrow V=u+at$
$V_{y}=2\space m/s+(-2\space m/s^{2})\times 0.5\space s=1\space m/s$
According to the Pythagorean theorem,
The magnitude of the puck's velocity $V=\sqrt {V_{x}^{y}+V_{y}^{y}}$
$V=\sqrt {(2\space m/s)^{2}+(1\space m/s)^{2}}=2.2\space m/s$
By using trigonometry, we can find the angle that the velocity makes with respect to the +x axis is,
$tan\theta=\frac{V_{y}}{V_{x}}=\frac{1\space m/s}{2\space m/s}=> \theta=tan^{-1}(0.5)=27^{\circ}$ above the +x axis
