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(C) Let the velocity of the ball be $\vec{v}_b = u \cos 	heta \hat{i} + u \sin 	heta \hat{j}$.Assuming the mirror is placed vertically in the $yz$-p

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Must be a straight line and horizontal

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(C) Let the velocity of the ball be $\vec{v}_b = u \cos 	heta \hat{i} + u \sin 	heta \hat{j}$.Assuming the mirror is placed vertically in the $yz$-plane,the velocity of the image of the ball $\vec{v}_i$ will be $\vec{v}_i = -u \cos 	heta \hat{i} + u \sin 	heta \hat{j}$.The velocity of the image with respect to the ball is given by $\vec{v}_{ib} = \vec{v}_i - \vec{v}_b$.Substituting the values: $\vec{v}_{ib} = (-u \cos 	heta \hat{i} + u \sin 	heta \hat{j}) - (u \cos 	heta \hat{i} + u \sin 	heta \hat{j})$.$\vec{v}_{ib} = -2u \cos 	heta \hat{i}$.Since the velocity is constant and directed along the negative $x$-axis,the motion is a straight line and horizontal.

$(a)$ Angle of reflection	$(i)$ $60^\circ$
$(b)$ Value of $\alpha$	(ii) $120^\circ$
$(c)$ Angle of deviation	(iii) $30^\circ$

$A$ ball is projected from the top of a table with an initial speed $u$ at an angle of inclination $\theta$ with the horizontal. What is the motion of the image of the ball with respect to the ball?

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