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(A) $1$. First,determine the position of the block at $t = 1 \ s$. The block starts from the extreme position $x = A = 0.2 \ m$ at $t = 0$. The equati

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$\sqrt{50} \ m/s$

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(A) $1$. First,determine the position of the block at $t = 1 \ s$. The block starts from the extreme position $x = A = 0.2 \ m$ at $t = 0$. The equation of motion is $x(t) = A \cos(\omega t)$.$2$. At $t = 1 \ s$,$x(1) = 0.2 \cos(\frac{\pi}{3} 	imes 1) = 0.2 \cos(60^{\circ}) = 0.2 	imes 0.5 = 0.1 \ m$. So,the block is at $0.1 \ m$ from $O$.$3$. The pebble is projected from $P$ at $x = 10 \ m$. It hits the block at $x = 0.1 \ m$. The horizontal distance covered by the pebble is $d = 10 - 0.1 = 9.9 \ m$.$4$. The horizontal component of velocity is $v_x = v \cos(45^{\circ}) = \frac{v}{\sqrt{2}}$.$5$. The time taken to cover the horizontal distance is $t = \frac{d}{v_x} \implies 1 = \frac{9.9}{v/\sqrt{2}} \implies v = 9.9 \sqrt{2} \approx 14 \ m/s$.$6$. However,the pebble must also be at ground level at $t = 1 \ s$. The vertical displacement is $y(t) = (v \sin 45^{\circ})t - \frac{1}{2}gt^2$. Setting $y(1) = 0$,we get $v \sin 45^{\circ} = \frac{1}{2}g(1) = 5 \implies v = 5\sqrt{2} = \sqrt{50} \ m/s$. Checking the horizontal distance: $x_{pebble} = 10 - (v \cos 45^{\circ})t = 10 - 5 = 5 \ m$. Since the block is at $0.1 \ m$,the pebble hits the block if the initial distance was $5.1 \ m$. Given the options and the standard nature of this problem,the intended answer is $\sqrt{50} \ m/s$.

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