The acceleration due to gravity on the planet | vedclass

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(A) The maximum height reached by a person jumping with an initial velocity $u$ is given by the formula $H_{\max} = \frac{u^2}{2g}$.Since the initial

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$18$

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(A) The maximum height reached by a person jumping with an initial velocity $u$ is given by the formula $H_{\max} = \frac{u^2}{2g}$.Since the initial velocity $u$ is the same for both jumps,we have $H_{\max} \propto \frac{1}{g}$.Let $g_A$ and $g_B$ be the acceleration due to gravity on planets $A$ and $B$ respectively. We are given $g_A = 9g_B$.Let $H_A = 2 \ m$ be the height on planet $A$ and $H_B$ be the height on planet $B$.Using the proportionality $H_A g_A = H_B g_B$,we get $H_B = H_A \left( \frac{g_A}{g_B} ight)$.Substituting the values,$H_B = 2 	imes 9 = 18 \ m$.

The acceleration due to gravity on the planet $A$ is $9$ times the acceleration due to gravity on planet $B$. $A$ man jumps to a height of $2 \ m$ on the surface of $A$. What is the height of jump by the same person on the planet $B$?

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