A ballistic pendulum of mass 5KG was struck by a bullet of mass 0.01KG. The centre of mass of the pendulum then rose by a vertical distance of 0.01m. Assuming that the bullet remained embedded in the penduum, calculate its initial speed (g=9.8m/s^2)

initial energy = final energy (principle of conservation of energy)
(kinetic E + potential E) initial = (kinetic E + potential E) final
0.5*m*Vi^2 + m*g*hi = 0.5*m*Vf^2 + m*g*hf

Take the ground as a reference, original position of the pendulum (hi =0) and the final position has h = 0.01 and Vf = 0 (stopped before going back to original position)

0.5*(5.01)*Vi^2 + 0 = 0 + 5.01*9.8*0.01
Vi^2 = 0.196
Vi = 0.442 m/sec

Thank you..