# 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.

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Equation of motion under gravitational force

`h=ut+(1/2)g t^2`

`` let boy jumps from statinary position i.e at rest at both surfaces earth and moon. So u=0 ,thus equation of motion now becomes

`h=(1/2)g t^2`

on earth h=2 metres

`2=(1/2)g t^2`

`(4/g)=t^2`

`t=sqrt(4/g)` (i)

on surface of the moon , let boy jumps for same time. and jumps to x metres.

`x=(1/2)(g/6)t^2`

`x=(1/6)((1/2)g t^2)`

`x=(1/6)2`

`x=1/3 metre.`

`` Here in problem so many informations were missing ,so it is difficult to give the desired result.