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(B) The swimmer performs horizontal projectile motion.Let $h = 10 \ m$ be the vertical height and $x = 2 \ m$ be the horizontal distance to clear.The

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$1.4$

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(B) The swimmer performs horizontal projectile motion.Let $h = 10 \ m$ be the vertical height and $x = 2 \ m$ be the horizontal distance to clear.The time taken to fall $10 \ m$ is given by $h = \frac{1}{2} g t^2$.Using $g = 9.8 \ m/s^2$,we get $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 	imes 10}{9.8}} = \sqrt{2.04} \approx 1.428 \ s$.To clear the ledge,the horizontal distance covered must be at least $2 \ m$.Using $x = v 	imes t$,we have $v = \frac{x}{t} = \frac{2}{1.428} \approx 1.4 \ m/s$.Thus,the minimum speed required is $1.4 \ m/s$.

$A$ swimmer dived off a cliff with a running horizontal leap. What must his minimum speed be just as he leaves the top of the cliff so that he will miss the edge at the bottom,given that the ledge at the bottom is $2 \ m$ wide and $10 \ m$ below the top of the cliff?

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