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 $$B$$ ?

Solution :
Applying conservation of total mechanical energy principle
$$\eqalign{ & \frac{1}{2}m{v^2} = m{g_A}{h_A} = m{g_B}{h_B} \cr & \Rightarrow {g_A}{h_A} = {g_B}{h_B} \cr & \Rightarrow {h_B} = \left( {\frac{{{g_A}}}{{{g_B}}}} \right){h_A} = 9 \times 2 = 18\,m \cr} $$