What initial speed must you give the second chestnut if they are both to reach the ground at the same time?

h = height
h(0) = initial height
v = initial velocity
g = acceleration due to gravity

Positive values for acceleration and velocity are downward.

h = h(0) – (1/2)gt^2 – vt

chestnut 1
To reach the ground:
0 = 12 – (1/2)gt^2 – 0
(1/2)gt^2 = 12
gt^2 = 24
t = sqrt(24/g) = 2 * sqrt(6)/sqrt(g)

To drop the first 2 meters { using w for the time waiting before throwing chestnut number 2}:
10 = 12 – (1/2)gw^2 – 0
(1/2)gw^2 = 2
gw^2 = 4
w = sqrt(4/g) = 2 / sqrt(g)

chestnut 2 to reach the ground
0 = 12 – (1/2)g(t – w)^2 – v(t – w)
substitute x for (t – w) = (2 * sqrt(6) – 2) / sqrt(g) = 2 * (sqrt(6) – 1) / sqrt(g)
0 = 12 – (1/2)gx^2 – vx
Solve for v
(1/2)gx^2 – 12 = -vx
((1/2)gx^2 – 12)/(-x) = v
12/x – (1/2)gx = v
12 * (sqrt(g) / (2 * sqrt(6) – 1)) – (1/2)g((sqrt(6) – 1) / sqrt(g)) = v