Momentum, $\quad p=m \cdot v$


$\begin{aligned}
& \Rightarrow \quad v=\left(\frac{p}{m}\right) \\
& \text { Kinetic energy, } \mathrm{KE}=\frac{1}{2} m v^2 \\
& =\frac{1}{2} m\left(\frac{p^2}{m^2}\right)=\frac{1}{2 m} p^2 \\
& \Rightarrow \quad \mathrm{KE} \propto p^2 \quad\left(\because \frac{1}{2 m}=\text { constant }\right) \\
&
\end{aligned}$
Hence, the graph between KE and $p^2$ will be linear as shown below

Now, kinetic energy KE $=\frac{1}{2} m v^2$
The velocity component at point $P$,
and
$\begin{aligned}
& v_y=(u \sin \theta-g t) \\
& v_x=u \cos \theta
\end{aligned}$
Resultant velocity at point $P$,
$\begin{aligned}
& \overrightarrow{\mathbf{v}}=v_y \hat{\mathbf{j}}+v_x \hat{\mathbf{i}} \\
& =(u \sin \theta-g t) \hat{\mathbf{j}}+u \cos \theta \hat{\mathbf{i}} \\
& |\overrightarrow{\mathbf{v}}|=\sqrt{(u \cos \theta)^2+(u \sin \theta-g t)^2} \\
& =\sqrt{u^2 \cos ^2 \theta+u^2 \sin ^2 \theta+g^2 t^2-2 u g t \sin \theta} \\
& \therefore=\sqrt{u^2\left(\cos ^2 \theta+\sin ^2 \theta\right)+g^2 t^2-2 u g t \sin \theta} \\
& \mathrm{KE}=\frac{1}{2} m\left(u^2+g^2 t^2-2 u g t \sin \theta\right) \\
& \Rightarrow \mathrm{KE} \propto t^2 \\
&
\end{aligned}$
Hence, graph will be parabolic with intercept on $y$-axis.
Hence, the graph between KE and $t$

Now, in case of height
$\mathrm{KE}=\frac{1}{2} m\left(v^2\right)$
and
$v^2=\left(u^2-2 g y\right)$
$\begin{aligned}
\therefore \quad \mathrm{KE} & =\frac{1}{2} m\left(u^2-2 g y\right) \\
\mathrm{KE} & =-m g \mathrm{y}+\frac{1}{2} m u^2
\end{aligned}$
Intercept on $y$-axis $=\frac{1}{2} m u^2$
Now,
$\begin{aligned}
\mathrm{KE} & =\frac{1}{2} m v^2 \\
\mathrm{KE} & =\frac{1}{2} m\left(\frac{x}{t}\right)^2
\end{aligned}$

Now,
$\begin{aligned}
\mathrm{KE} & =\frac{1}{2} m v^2 \\
\mathrm{KE} & =\frac{1}{2} m\left(\frac{x}{t}\right)^2
\end{aligned}$

$\mathrm{KE} \propto x^2$. Thus graph between KE and $x$ will be parabolic.