A gun is fired vertically into a 1.75 kg block of wood at rest directly above it. If the bullet has a mass of 21.0 g and a speed of 190 m/s?

Take g = 9.81 m/s²
Mass of bullet = 21.0 g = 0.0210 kg

Loss in K.E. of the bullet = Gain in P.E. of the wood and the bullet
(1/2) × m × v² = (m + M) × g × h
(1/2) × 0.0210 × 190² = (0.0210 + 1.75) × 9.81 × h
h = (1/2) × 0.0210 × 190² / [(0.0210 + 1.75) × 9.81] m = 21.8 m

Height that the block of wood rises = 21.8 m

From the COM law we have mU = (M + m)v and then from COE law we have h = KE/(M + m)g = .5 (M + m)v^2/(M + m)g = v^2/2g = (mU/(M + m))^2/2g = (.021*190/(1.75 + .021))^2/(2*9.8) = .26 m. ANS.

Energy is NOT conserved, much is lost by the bullet penetrating the wood.

Momentum is conserved. Momentum (P) of the bullet is
P = mV = 0.021 kg x 190 m/s = 3.99 kgm/s

that is imparted to the block which now weighs 1.75+0.021 kg
mV = 3.99
V = 3.99 / (1.75+0.021) = 2.253 m/s

now eq. of motion
h = v²/2g = 2.253²/2•9.8 = 0.259 meters

Given quantities:
m = 0.021 kg
M = 1.75 kg
v = 190 m/s
g = 9.80 m/s²
Unknown quantities:
V = ?
h = ?

From conservation of momentum, we set up an equation that describes momenta before and after
the collision, which is inelastic in this case.
mv + M*0 = (m+M)V
=> V = mv/(m+M)

Now using conservation of energy, where the kinetic energy of the m+M system will be converted into
potential energy
Kinetic energy: K = ½*mass*speed²
Potential energy: U = mass*gravitational acceleration*height

==> ½(m+M)V² = (m+M)gh
==> ½(m+M)(mv/(m+M))² = (m+M)gh
==> ½(mv/(m+M))² = gh
==> h = (mv/(m+M))² / 2g

Insert the quantites m, v, M, g and calculate:

==> h = 0.259 m
Which should be rounded to 0,26m considering the significant figures in 190m/s.