Momentum, $p=m \cdot v \Rightarrow v=\left(\frac{p}{m}\right)$


Kinetic energy, KE
$$
\begin{array}{l}
=\frac{1}{2} m v^{2}=\frac{1}{2} m\left(\frac{p^{2}}{m^{2}}\right)=\frac{1}{2 m} p^{2} \\
\text { or, } \mathrm{KE} \propto p^{2} \quad\left(\because \frac{1}{2 m}=\text { constant }\right)
\end{array}
$$
Hence, the graph between $\mathrm{KE}$ and $p^{2}$ will be linear.
Now, kinetic energy $\mathrm{KE}=\frac{1}{2} m v^{2}$
The velocity component at point $P$,
$v_{y}=(u \sin \theta-g t)$ and $v_{x}=u \cos \theta$
Resultant velocity at point $P$
$$
\begin{aligned}
\vec{v} &=v_{y} \hat{j}+v_{x} \hat{i} \\
&=(u \sin \theta-g t) \hat{j}+u \cos \theta \hat{i} \\
&|\vec{v}|=\sqrt{(u \cos \theta)^{2}+(u \sin \theta-g t)^{2}} \\
=\sqrt{u^{2} \cos ^{2} \theta+u^{2} \sin ^{2} \theta+g^{2} t^{2}-2 u g t \sin \theta} \\
&=\sqrt{u^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+g^{2} t^{2}-2 u g t \sin \theta}
\end{aligned}
$$
$$
\therefore \mathrm{KE}=\frac{1}{2} m\left(u^{2}+g^{2} t^{2}-2 u g t \sin \theta\right)ii/
$$
i. e., $\mathrm{KE} \propto t^{2}$
Hence, graph will be parabolic intercept on $y$-axis.
Hence, the graph between $\mathrm{KE}$ and $t$. Now, in case of height
$$
\begin{array}{l}
\mathrm{KE}=\frac{1}{2} m\left(v^{2}\right) \text { and } v^{2}=\left(u^{2}-2 g y\right) \\
\therefore \mathrm{KE}=\frac{1}{2} m\left(u^{2}-2 g y\right) \\
\mathrm{KE}=-m g y+\frac{1}{2} m u^{2}
\end{array}
$$
Intercept on $y$-axis $=\frac{1}{2} m u^{2}$


$$
\mathrm{KE}=\frac{1}{2} m\left(\frac{x}{t}\right)^{2}
$$
i.e., $\mathrm{KE} \propto x^{2}$. Thus graph between $\mathrm{KE}$ and $x$ will be parabolic.