A particle is released from height S from the surface of the Earth....

$∵$ At Every $pt$.

$KE + PE =$ constant

so,

$( KE{)}_{A} + ( PE{)}_{A} = ( KE{)}_{B} + ( PE{)}_{B}$

$0 + mgS = 3 ( PE{)}_{B} + ( PE{)}_{B}$

$mgS = 4mgh$

$h = \dfrac{S}{4}$

Now,

$KE$ at $pt . B -$

$∵$ At $pt . B -$

$KE = 3PE$

$\dfrac{1}{2}m{v}^{2} = 3\dfrac{mgs}{4}$

$v = \sqrt{\dfrac{3}{2}gS}$