According to the equation of motion under gravity v^{2} − u^{2} = 2gs

Where,

u = Initial velocity of the stone = 0 m/s

v = Final velocity of the stone

s = Height of the stone = 19.6 m

g = Acceleration due to gravity = 9.8 ms^{−2 }

∴ v ^{2} − 0^{2} = 2 × 9.8 × 19.6

⇒ v ^{2} = 2 × 9.8 × 19.6 = (19.6)^{2}

⇒ v = 19.6 ms^{−1 }

Hence, the velocity of the stone just before touching the ground is 19.6 ms^{−1} .