From 18 text{ m} height above the ground,a ba | vedclass

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(A) Let the ball fall a distance $h'$ from the top.The velocity $v$ of the ball after falling a distance $h'$ is given by $v = \sqrt{2gh'}$.The magnit

Vedclass · Accepted Answer

$13$

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(A) Let the ball fall a distance $h'$ from the top.The velocity $v$ of the ball after falling a distance $h'$ is given by $v = \sqrt{2gh'}$.The magnitude of acceleration due to gravity is $a = g = 10 	ext{ m/s}^2$.According to the problem,the magnitude of velocity equals the magnitude of acceleration: $|v| = |a|$.Substituting the values,we get $\sqrt{2gh'} = g$.Squaring both sides,we get $2gh' = g^2$.Dividing by $g$,we get $2h' = g$.Given $g = 10 	ext{ m/s}^2$,we have $2h' = 10$,which gives $h' = 5 	ext{ m}$.The height above the ground $H$ is the total height minus the distance fallen: $H = 18 	ext{ m} - 5 	ext{ m} = 13 	ext{ m}$.

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