The acceleration due to gravity on the planet | vedclass

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(D) Given that the acceleration due to gravity on planet $A$ is $9$ times that on planet $B$:$g_{A} = 9g_{B}$ $(i)$Using the third equation of motion,

Vedclass · Accepted Answer

$18 \,m$

Vedclass · Answer

(D) Given that the acceleration due to gravity on planet $A$ is $9$ times that on planet $B$:$g_{A} = 9g_{B}$ $(i)$Using the third equation of motion, $v^2 = u^2 + 2gh$. Since the initial velocity $u$ is the same for the jump on both planets and the final velocity $v$ at the maximum height is $0$, we have $u^2 = 2gh$, or $h = \frac{u^2}{2g}$.Since $u$ is constant for the same person,$h \propto \frac{1}{g}$Therefore, $\frac{h_{B}}{h_{A}} = \frac{g_{A}}{g_{B}}$Substituting the given values:$\frac{h_{B}}{2} = \frac{9g_{B}}{g_{B}} = 9$$h_{B} = 9 	imes 2 = 18 \,m$.

Column-$I$	Column-$II$
$(1)$ Maximum value of acceleration due to gravity $g$	$(a)$ At the center of the Earth
$(2)$ Minimum value of acceleration due to gravity $g$	$(b)$ At the poles
$(3)$ Zero value of acceleration due to gravity $g$	$(c)$ At the equator

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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Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Maximum value of acceleration due to gravity $g$ $(a)$ At the center of the Earth

$(2)$ Minimum value of acceleration due to gravity $g$ $(b)$ At the poles

$(3)$ Zero value of acceleration due to gravity $g$ $(c)$ At the equator

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