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 `2m` on the surface of `A`. What is the height of jump by the same person on the planet `B`?

Options

A. 6 m
B. `2//3` m
C. `2//9` m
D. 18 m

Answer

D

Explanation

It is given that acceleratini due to gravity on planet A is 9 times the acceleration due to gravity on planet B i.e
`g_(A)=9g`
From third equatin of motion `v^(2)=2gh`
At planet A `h_(A)=(V^(2))/(23g_(A))`
At planet B `h_(B)=(v^(2))/(2g_(B))`
for safe jump `v_(1)=v_(2)`
Dividing Eq ii by Eq iii we have
`(h_(A))/(H_(B))=(g_(B))/(g_(A))`
`rarr (h_(A))/(h_(B))=1/9`
`rarr h_(B)=9h_(A)=9xx2=18m`