Answer
$88.42N , \space Direction-76.7^{\circ} from\space x+\space axis$
Work Step by Step
Please see the image first.
Let's take,
Mass of the puck (m) $=166\times10^{-3}\space kg$
Time of force acted (t) $=112\times10^{-3}\space s$
Force applied $= F$
x- direction component of force $=F_{x}$
y- direction component of force $=F_{y}$
In here we have to apply Newton's second law in both x,y directions seperately as follows.
$\rightarrow F=ma$
Let's plug known values into this equation.
$F_{x}=166\times10^{-3}\space kg\times\frac{(82.1cos45^{\circ}-44.3)\space m/s}{112\times10^{-3}s}$
$F_{x}=20.38\space N$
$\uparrow F=ma$
Let's plug known values into this equation.
$F_{y}=166\times10^{-3}\space kg\times\frac{(82.1sin45^{\circ}-0)\space m/s}{112\times10^{-3}s}$
$F_{y}=86.04\space N$
$F=\sqrt{F_{x}^{2}+F_{y}^{2}}$
$F=\sqrt {(20.38\space N)^{2}+(86.04\space N)^{2}}=88.42\space N$
$tan\alpha=\frac{F_{y}}{F_{x}}= \frac{86.04}{20.38}= 4.2$
$\alpha=tan^{-1}(4.2)=76.7^{\circ}$
