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(A) $1$. Kinetic energy $(KE)$ and momentum $(p)$: $KE = \frac{p^2}{2m}$. Since $m$ is constant,$KE \propto p^2$. The graph of $KE$ versus $p^2$ is a

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Graph $(A)$

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(A) $1$. Kinetic energy $(KE)$ and momentum $(p)$: $KE = \frac{p^2}{2m}$. Since $m$ is constant,$KE \propto p^2$. The graph of $KE$ versus $p^2$ is a straight line passing through the origin. $2$. Kinetic energy $(KE)$ and time $(t)$: $v_x = u \cos 	heta$ and $v_y = u \sin 	heta - gt$. $KE = \frac{1}{2}m(v_x^2 + v_y^2) = \frac{1}{2}m(u^2 \cos^2 	heta + (u \sin 	heta - gt)^2)$. This is a quadratic equation in $t$,representing a parabola. $3$. Kinetic energy $(KE)$ and vertical displacement $(y)$: Using $v_y^2 = u_y^2 - 2gy$,we have $KE = \frac{1}{2}m(v_x^2 + v_y^2) = \frac{1}{2}m(u^2 \cos^2 	heta + u^2 \sin^2 	heta - 2gy) = \frac{1}{2}mu^2 - mgy$. This is a linear equation of the form $KE = -mgy + C$,which represents a straight line with a negative slope. $4$. Analysis of Graph $(A)$: The graph $(A)$ shows a $V$-shaped curve for $KE$ versus $y$. However,the relationship derived is $KE = \frac{1}{2}mu^2 - mgy$,which is a linear relationship,not a $V$-shaped curve. Therefore,graph $(A)$ does not represent the variation of $KE$ correctly.

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