The acceleration due to gravity on planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2m on the surface A. What is the height of the jump by the same person on the planet B?

It is given that acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B i.e.,

g_A​=9g_B​ ...(i)

From the third equation of motion 

v²=2gh ...(ii)

At planet A, h_A​=$\frac{v^2}{2g_A}$​​ ...(ii)

At planet B, h_B​=$\frac{v^2}{2g_B}$​​ ...(iii)

Dividing Eq. (ii) by Eq. (iii) we have 

$\frac{h_A}{h_B}=\frac{g_B}{g_A}$

From Eq. (i), gA​=9gB​

∴$\frac{h_A}{h_B}=\frac{g_A}{9g_B}$ = $\frac{1}{9}$

or h_B​=9h_A​=9×2

                 =18 m(∴h_A​=2m)