We have, $v_{x}=v_{0} \cos \alpha, v_{y}=v_{0} \sin \alpha-g t$
As $\vec{v} \uparrow \uparrow \hat{u}_{t}$ all the moments of time.
Thus $v^{2}=v_{t}^{2}-2 g t v_{0} \sin \alpha+g^{2} t^{2}$
Now, $\quad w_{t}=\frac{d v_{t}}{d t}=\frac{1}{2 v_{t}} \frac{d}{d t}\left(v_{t}^{2}\right)=\frac{1}{v_{t}}\left(g^{2} t-g v_{0} \sin \alpha\right)$
$=-\frac{g}{v_{t}}\left(v_{0} \sin \alpha-g t\right)=-g \frac{v_{y}}{v_{t}}$
Hence $\quad\left|w_{t}\right|=g \frac{\left|v_{y}\right|}{v}$
Now $w_{n}=\sqrt{w^{2}-w_{t}^{2}}=\sqrt{g^{2}-g^{2} \frac{v_{y}^{2}}{v_{t}^{2}}}$
or $\quad w_{n}=g \frac{v_{x}}{v_{t}}\left(\right.$ where $\left.v_{x}=\sqrt{v_{t}^{2}-v_{y}^{2}}\right)$
As $\quad \vec{v} \uparrow \uparrow \hat{v}_{t}$, during time of motion
$w_{v}=w_{t}=-g \frac{v_{y}}{v}$
On the basis of obtained expressions or facts the sought plots can be drawn as shown in the figure of answer sheet.