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(A) The kinetic energy $KE$ is given by $KE = \frac{p^2}{2m}$. Since $m$ is constant, $KE \propto p^2$. Thus, the graph between $KE$ and $p^2$ is a st

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Graph $(A)$

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(A) The kinetic energy $KE$ is given by $KE = \frac{p^2}{2m}$. Since $m$ is constant, $KE \propto p^2$. Thus, the graph between $KE$ and $p^2$ is a straight line passing through the origin. This matches graph $(D)$.For vertical displacement $y$, using $v_y^2 = (u \sin 	heta)^2 - 2gy$, the kinetic energy is $KE = \frac{1}{2}m(v_x^2 + v_y^2) = \frac{1}{2}m(u^2 \cos^2 	heta + u^2 \sin^2 	heta - 2gy) = \frac{1}{2}mu^2 - mgy$. This is a linear equation of the form $KE = -mgy + C$, which represents a straight line with a negative slope. Graph $(A)$ shows a $V$-shape, which is incorrect for this linear relationship.For time $t$, $KE = \frac{1}{2}m(v_x^2 + v_y^2) = \frac{1}{2}m(u^2 \cos^2 	heta + (u \sin 	heta - gt)^2)$. This is a quadratic equation in $t$ $(KE \propto t^2)$, representing a parabola. Graph $(B)$ shows a parabolic variation.For horizontal displacement $x$, $x = (u \cos 	heta)t \Rightarrow t = \frac{x}{u \cos 	heta}$. Substituting this into the $KE$ expression results in a quadratic relationship $KE \propto x^2$, which is also parabolic. Graph $(C)$ shows a parabolic variation.Therefore, graph $(A)$ does not correctly represent the variation of $KE$ with $y$.

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