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(C) $1$. First,calculate the velocity $v$ of the ball as it leaves the spring gun using the principle of conservation of mechanical energy:$\frac{1}{2

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(C) $1$. First,calculate the velocity $v$ of the ball as it leaves the spring gun using the principle of conservation of mechanical energy:$\frac{1}{2} k x^2 = \frac{1}{2} m v^2$Given $k = 100\, 	ext{N/m}$,$x = 0.05\, 	ext{m}$,and $m = 100\, 	ext{g} = 0.1\, 	ext{kg}$.$100 	imes (0.05)^2 = 0.1 	imes v^2$$100 	imes 0.0025 = 0.1 	imes v^2$$0.25 = 0.1 	imes v^2$$v^2 = 2.5$$v = \sqrt{2.5}\, 	ext{m/s}$.$2$. Next,calculate the time $t$ taken by the ball to reach the ground from a height $h = 2\, 	ext{m}$ using the equation of motion for horizontal projection:$h = \frac{1}{2} g t^2$$2 = \frac{1}{2} 	imes 10 	imes t^2$$2 = 5 t^2$$t^2 = 0.4$$t = \sqrt{0.4}\, 	ext{s}$.$3$. Finally,calculate the horizontal distance $d$:$d = v 	imes t$$d = \sqrt{2.5} 	imes \sqrt{0.4}$$d = \sqrt{2.5 	imes 0.4}$$d = \sqrt{1} = 1\, 	ext{m}$.Thus,the value of $d$ is $1\, 	ext{m}$.

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