A boy jumps a distance of 2m on the surface of the earth. What distance will he jump on the surface of the moon where g is 1/6th of the value on the surface of the earth?

Answers (1)

 \begin{array}{l} \text { Given } \frac{g_{m}}{g_{e}}=\frac{1}{6} \\ \Rightarrow g_{e}=6 g_{m} \end{array}

so using
$v^{2}=u^{2}-2 g s \quad$

where

g= acceleration due to gravity

s= total height

and v=0

then we get

for earth

\begin{array}{l} 0=u^{2}-2 g s \\ \Rightarrow s=\frac{u^{2}}{2 g}\\ For \ earth \ 2=\frac{u^{2}}{2 g_{e}}\\ for \ moon \ \ h=\frac{u^{2}}{2 g_{m}} \end{array}

So using g_{e}=6 g_{m}

we get

h=12 m

so the boy can jump height of 12 m on moon's surface
 

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