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(A) In projectile motion,the acceleration is constant and directed vertically downwards (gravity). Let the velocity vector be $\vec{v} = v_x \hat{i} +

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(A) In projectile motion,the acceleration is constant and directed vertically downwards (gravity). Let the velocity vector be $\vec{v} = v_x \hat{i} + v_y \hat{j}$ and the acceleration vector be $\vec{a} = -g \hat{j}$. The angle $	heta$ between $\vec{v}$ and $\vec{a}$ is given by $\cos 	heta = \frac{\vec{v} \cdot \vec{a}}{|\vec{v}| |\vec{a}|} = \frac{-g v_y}{\sqrt{v_x^2 + v_y^2} \cdot g} = \frac{-v_y}{\sqrt{v_x^2 + v_y^2}}$. As the projectile moves,$v_y$ changes from positive (upward) to negative (downward). When $v_y > 0$,$\cos 	heta$ is negative,meaning $	heta$ is obtuse $(> 90^{\circ})$. When $v_y As $v_y$ becomes more negative,$\cos 	heta$ increases,meaning $	heta$ decreases. Thus,the angle is acute throughout the downward journey and reaches its minimum value at the point of impact (where $v_y$ is most negative). However,in the context of standard physics problems,the angle is considered acute during the entire descent. The question asks for the point where it is minimum and acute; this occurs at the final point of the trajectory.

For a projectile motion,the angle between the velocity and acceleration is minimum and acute at

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