An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a 1 m high table. The bullet remains in the block, and after the impact the block lands 2.00 m from the bottom of the table. Determine the initial speed of the bullet

So the block fell through a height of 1 m in time t, where

1.0 = (1/2) gt²

or

t=√ (2/9.81)

=0.452 s

During this time, the block has travelled horizontally 2m, or

horiz. velocity

=2/0.452

=4.43 m/s

This velocity, v, is the velocity of the block after impact, given by the equation of momentum before and after impact, namely

m1u1+m2u2 = (m1+m2) v

or

v = (m1u1+m2u2) / (m1+m2)

where

m1=8g

m2=250g

u1 = to be determined

u2=0 (block)

Solve for u1 (velocity of bullet)