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(A) Let $W_f$ be the work done by air resistance during the upward journey. Since the work done by air resistance is the same for both paths,the work

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$8$

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(A) Let $W_f$ be the work done by air resistance during the upward journey. Since the work done by air resistance is the same for both paths,the work done during the downward journey is also $W_f$.Applying the Work-Energy Theorem for the upward journey from the point of projection to the maximum height $H$:$W_g + W_f = \Delta KE$$-mgH - W_f = 0 - \frac{1}{2}m(16)^2$$mgH + W_f = 128m$  --- $(1)$Applying the Work-Energy Theorem for the downward journey from the maximum height $H$ back to the point of projection:$W_g + W_f = \Delta KE$$mgH - W_f = \frac{1}{2}m(8)^2 - 0$$mgH - W_f = 32m$  --- $(2)$Adding equations $(1)$ and $(2)$:$2mgH = 160m$$2gH = 160$$2(10)H = 160$$20H = 160$$H = 8 \ m$.

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