Saved Bookmarks
| 1. |
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` ?A. 6mB. `(2)/(3)m`C. `2//9`mD. 18m |
|
Answer» Correct Answer - D It is given that, acceleration due to gravity on planet A is 9 times the acceleration due to gravity on planet ie, `g_(A)=9_(g_(B)).....(i)` For third equation of motion `v^(2)=2gh` At plaenet A, `h_(A)=(v^(2))/(2g_(A)),......(ii)` At plaenet B, `h_(B)=(v^(2))/(2g_(B))......(iii)` Dividing Eq. (ii) by Eq. (iii), we have `(h_(A))/(h_(B))=(g_(B))/(g_(A))` From Eq. (i) `g_(A)=9g_(B)` `therefore (h_(A))/(h_(B))=(g_(B))/(9g_(B))=(1)/(9)` or `h_(B)=9h_(A)=9xx2=18m (therefore h_(A)=2m)` |
|