The mass of the system block + bullet = 3.75 kg
The initial speed of the system bullet + block is given by
Vf² = Vi² +2*g*d...........
Vf = 0 (Final speed equals zero)
d is the height traveled in the vertical component (0.5m)
Vi = sqrt(-2*(-9.8[m/s²])* (0.5m)) = 3.13 m/s
The momentum is conserved, so
m*V1 (before impact) = M*Vi (after impact)
0.75kg*V1 = 3.75kg*(3.13m/s)
>> V1 = 15.65 [m/s] (Initial speed of the bullet)
