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(B) Let the distance fallen be $s$. For a particle starting from rest,the velocity $v_i$ and time $t_i$ taken on a planet with acceleration $a_i$ are

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$v = 4gt$

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(B) Let the distance fallen be $s$. For a particle starting from rest,the velocity $v_i$ and time $t_i$ taken on a planet with acceleration $a_i$ are given by $v_i = \sqrt{2a_is}$ and $t_i = \sqrt{2s/a_i}$.Given $a_1 = 2g$ and $a_2 = 8g$,we have $v_1 = \sqrt{2(2g)s} = 2\sqrt{gs}$ and $v_2 = \sqrt{2(8g)s} = 4\sqrt{gs}$.Similarly,$t_1 = \sqrt{2s/2g} = \sqrt{s/g}$ and $t_2 = \sqrt{2s/8g} = 0.5\sqrt{s/g}$.The change in velocity is $v = v_2 - v_1 = 4\sqrt{gs} - 2\sqrt{gs} = 2\sqrt{gs}$.The change in time is $t = t_1 - t_2 = \sqrt{s/g} - 0.5\sqrt{s/g} = 0.5\sqrt{s/g}$.From the second equation,$\sqrt{s/g} = 2t$. Substituting this into the velocity equation: $v = 2\sqrt{g} \cdot \sqrt{s} = 2\sqrt{g} \cdot (2t\sqrt{g}) = 4gt$.

If a particle takes $t$ seconds less and acquires a velocity of $v \ ms^{-1}$ more in falling through the same distance (starting from rest) on two planets where the accelerations due to gravity are $2g$ and $8g$ respectively,then $v=$

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