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(A) The velocity components of a projectile at time $t$ are $v_x = u \cos 	heta$ and $v_y = u \sin 	heta - gt$.The kinetic energy $K$ is given by $K

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(A) The velocity components of a projectile at time $t$ are $v_x = u \cos 	heta$ and $v_y = u \sin 	heta - gt$.The kinetic energy $K$ is given by $K = \frac{1}{2} m v^2 = \frac{1}{2} m (v_x^2 + v_y^2)$.Substituting the components,$K = \frac{1}{2} m [(u \cos 	heta)^2 + (u \sin 	heta - gt)^2]$.Expanding this,$K = \frac{1}{2} m [u^2 \cos^2 	heta + u^2 \sin^2 	heta + g^2 t^2 - 2 u g t \sin 	heta] = \frac{1}{2} m [u^2 + g^2 t^2 - 2 u g t \sin 	heta]$.This is a quadratic equation in $t$ of the form $K = at^2 + bt + c$,which represents a parabola opening upwards.At $t = 0$,$K = \frac{1}{2} m u^2$. At the maximum height $(t = \frac{u \sin 	heta}{g})$,the vertical velocity is zero,so $K$ is minimum but non-zero $(K_{min} = \frac{1}{2} m (u \cos 	heta)^2)$.Thus,the graph is a parabola that decreases until the maximum height and then increases,which matches the shape in Graph $A$.

$A$ ball is projected from the ground at an angle of $\theta$ from the horizontal. The graph of kinetic energy $(K)$ versus time $(t)$ will be:

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