A small ball is projected up a smooth incline | vedclass

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(D) Let $	heta$ be the angle of inclination of the plane.The acceleration of the ball along the plane is $g \sin 	heta$ downwards,perpendicular to t

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(D) Let $	heta$ be the angle of inclination of the plane.The acceleration of the ball along the plane is $g \sin 	heta$ downwards,perpendicular to the bottom edge.The initial velocity component perpendicular to the bottom edge is $u_{\perp} = u \sin 30^{\circ} = 10 \sin 30^{\circ} = 5 \ m/s$.The acceleration component perpendicular to the bottom edge is $a_{\perp} = g \sin 	heta$.Since the ball returns to the edge,the displacement perpendicular to the edge is zero.Using $s = ut + \frac{1}{2}at^2$ in the direction perpendicular to the edge:$0 = (u \sin 30^{\circ})T - \frac{1}{2}(g \sin 	heta)T^2$$T = \frac{2 u \sin 30^{\circ}}{g \sin 	heta}$Given $T = 2 \ s$,$u = 10 \ m/s$,and $g = 10 \ m/s^2$:$2 = \frac{2 	imes 10 	imes \sin 30^{\circ}}{10 \sin 	heta}$$2 = \frac{2 	imes 0.5}{\sin 	heta}$$2 = \frac{1}{\sin 	heta}$$\sin 	heta = 0.5$$	heta = 30^{\circ}$.

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