A particle is projected up from a point at an angle $\mathrm{\theta }$ with the horizontal direction. At any time 't', if p is the linear momentum, y is the vertical displacement, 'x' is horizontal displacement, the graph among the following which does not represent the variation of kinetic energy 'k' of the particle is
 

Momentum, $\mathrm{p}=\mathrm{m}·\mathrm{v}\Rightarrow \mathrm{v}=\left(\frac{\mathrm{p}}{\mathrm{m}}\right)$
Kinetic energy,$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$

$\mathrm{KE}=\frac{{\mathrm{P}}^2}{2\mathrm{m}} ; \mathrm{KE}\propto {\mathrm{P}}^2$

$\therefore$graph is straight line passing through origin.

The velocity component at point $P, v_{\mathrm{y}}=(\mathrm{usin}\theta -\mathrm{gt})$ And ${\mathrm{v}}_{\mathrm{x}}=\mathrm{ucos}\theta$
Resultant velocity at point $\mathrm{P}, \overrightarrow{\mathrm{v}}={\mathrm{v}}_{\mathrm{y}}\hat{\mathrm{j}}+{\mathrm{v}}_{\mathrm{x}}\hat{\mathrm{i}}=(\mathrm{usin\theta }-\mathrm{gt})\hat{\mathrm{j}}+\mathrm{ucos}\hat{\mathrm{i}}$
$\left|\overrightarrow{\mathrm{v}}\right|=\sqrt{(\mathrm{ucos}\theta {)}^2+(\mathrm{usin}\theta -\mathrm{gt}{)}^2}\Rightarrow \mathrm{KE}\propto {\mathrm{t}}^2$
Hence, graph will be parabolic with intercept on y-axis. Hence, the graph between KE and t is,

$\text{ Now, in case of height, }\mathrm{KE}=-\mathrm{mgy}+\frac{1}{2}{\mathrm{mu}}^2$

graph should be straight line with decreasing slope.

putting the value of y in above equation, $KE \propto x^2$