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

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

The velocity component at point P,     ${\mathrm{v}}_{\mathrm{y}}=\left(\mathrm{usin\theta }-\mathrm{gt}\right)$And    ${\mathrm{v}}_{\mathrm{x}}=\mathrm{ucos\theta }$

Resultant velocity at point P,      $\overrightarrow{\mathrm{v}}={\mathrm{v}}_{\mathrm{y}}\widehat{\mathrm{j}}+{\mathrm{v}}_{\mathrm{x}}\widehat{\mathrm{i}}$ $=\left(\mathrm{usin\theta }-\mathrm{gt}\right)\widehat{\mathrm{j}}+\mathrm{ucos}\widehat{\mathrm{i}}$  

$\left|\overrightarrow{\mathrm{v}}\right|=\sqrt{{\left(\mathrm{ucos\theta }\right)}^2+{\left(\mathrm{usin\theta }-\mathrm{gt}\right)}^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

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