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(C) Let the initial velocity be $u$ at an angle $	heta$ with the horizontal. The horizontal component of velocity is $v_x = u \cos 	heta$,which rema

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(C) Let the initial velocity be $u$ at an angle $	heta$ with the horizontal. The horizontal component of velocity is $v_x = u \cos 	heta$,which remains constant throughout the motion. The vertical component of velocity is $v_y = u \sin 	heta - gt$. The horizontal displacement is $x = (u \cos 	heta)t$,so $t = \frac{x}{u \cos 	heta}$. Substituting $t$ into $v_y$,we get $v_y = u \sin 	heta - g \left( \frac{x}{u \cos 	heta} ight)$. The kinetic energy is $KE = \frac{1}{2} m (v_x^2 + v_y^2) = \frac{1}{2} m \left( (u \cos 	heta)^2 + (u \sin 	heta - \frac{gx}{u \cos 	heta})^2 ight)$. This is a quadratic equation in $x$ of the form $KE = ax^2 + bx + c$,where $a > 0$. Thus,the graph of $KE$ versus $x$ is a parabola opening upwards. At the highest point,$v_y = 0$,so $KE$ is minimum but not zero,as $v_x$ is non-zero. Graph $C$ represents this behavior.

$A$ ball is thrown up with a certain velocity at an angle $\theta$ to the horizontal. The kinetic energy $KE$ of the ball varies with horizontal displacement $x$ as

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