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(A) Let the distance fallen be $h$. The time taken to fall through distance $h$ under gravity $g'$ is $T = \sqrt{\frac{2h}{g'}}$ and the final velocit

Vedclass · Accepted Answer

$v=4gt$

Vedclass · Answer

(A) Let the distance fallen be $h$. The time taken to fall through distance $h$ under gravity $g'$ is $T = \sqrt{\frac{2h}{g'}}$ and the final velocity is $V = \sqrt{2g'h}$.For the first planet $(g_1 = 2g)$: $T_1 = \sqrt{\frac{2h}{2g}} = \sqrt{\frac{h}{g}}$ and $V_1 = \sqrt{2(2g)h} = 2\sqrt{gh}$.For the second planet $(g_2 = 8g)$: $T_2 = \sqrt{\frac{2h}{8g}} = \frac{1}{2}\sqrt{\frac{h}{g}}$ and $V_2 = \sqrt{2(8g)h} = 4\sqrt{gh}$.Given $t = T_1 - T_2 = \sqrt{\frac{h}{g}} - \frac{1}{2}\sqrt{\frac{h}{g}} = \frac{1}{2}\sqrt{\frac{h}{g}}$.Given $v = V_2 - V_1 = 4\sqrt{gh} - 2\sqrt{gh} = 2\sqrt{gh}$.From the time equation,$\sqrt{h} = 2t\sqrt{g}$.Substituting this into the velocity equation: $v = 2\sqrt{g}(2t\sqrt{g}) = 4gt$.

If a particle takes $t$ seconds less and acquires a velocity of $v \text{ m/s}$ more in falling through the same distance on two planets,where the accelerations due to gravity are $2g$ and $8g$ respectively,then:

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