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(C) The initial velocity of the ball is $u = 5 \,m \,s^{-1}$ at an angle of $30^{\circ}$ with the horizontal.We resolve this velocity into horizontal

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$\sqrt{19}$

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(C) The initial velocity of the ball is $u = 5 \,m \,s^{-1}$ at an angle of $30^{\circ}$ with the horizontal.We resolve this velocity into horizontal and vertical components:Horizontal component, $v_{1x} = u \cos 30^{\circ} = 5 	imes \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \,m \,s^{-1}$.Vertical component, $v_{1y} = u \sin 30^{\circ} = 5 	imes \frac{1}{2} = 2.5 \,m \,s^{-1}$.During the collision with the ground, the horizontal component of velocity remains unchanged because there is no impulse along the horizontal direction.$v_{2x} = v_{1x} = \frac{5\sqrt{3}}{2} \,m \,s^{-1}$.The vertical component changes according to the coefficient of restitution $e = 0.2$:$v_{2y} = e 	imes v_{1y} = 0.2 	imes 2.5 = 0.5 \,m \,s^{-1}$.The final speed of the ball $v_f$ is the magnitude of the resultant velocity vector:$v_f = \sqrt{v_{2x}^2 + v_{2y}^2} = \sqrt{\left(\frac{5\sqrt{3}}{2}ight)^2 + (0.5)^2}$$v_f = \sqrt{\frac{25 	imes 3}{4} + 0.25} = \sqrt{\frac{75}{4} + \frac{1}{4}} = \sqrt{\frac{76}{4}} = \sqrt{19} \,m \,s^{-1}$.

$A$ cricket ball hits the ground with a speed of $5 \,m \,s^{-1}$ making an angle of $30^{\circ}$ with the horizontal. If the coefficient of restitution is $0.2$, the speed of the reflected ball in $m \,s^{-1}$ is

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