Explanation

(d) It is given that, acceleration due to gravity on planet $A$ is 9 times the acceleration due to gravity on planet $B\ ie,$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ g_{A} = 9g_{B}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
From third equation of motion
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ v^{2} = 2gh$
$At\ planet\ A,\ {\ h}_{A} = \frac{v^{2}}{2g_{A}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
$At\ planet\ B,\ \ h_{B} = \frac{v^{2}}{2g_{B}}\ \ \ \ \ \ \ldots(iii)$
Dividing Eq. (ii) by Eq. (iii), we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{h_{A}}{h_{B}} = \frac{g_{B}}{g_{A}}$
From Eq. (i), $g_{A} = 9g_{B}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{h_{A}}{h_{B}} = \frac{g_{B}}{9g_{B}} = \frac{1}{9}$
$or\ \ \ \ \ \ \ \ h_{B} = 9h_{A} = 9 \times 2 = 18\ m\ \ \ \ \ \ \ \ (\therefore h_{A} = 2m)\ \ \ $