posted by having trouble .
An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height (Fig. P6.26). 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
During this time, the block has travelled horizontally 2m, or
This velocity, v, is the velocity of the block after impact, given by the equation of momentum before and after impact, namely
u1= to be determined
Solve for u1 (velocity of bullet)
do the end first
mass m (does not matter what m is) lands 2 meters from the table
how long did it fall?
h = 0 at end
0 = 1 - (1/2)(9.8) t^2
2 = 9.8 t^2
t = .45 seconds to fall to floor off table = time in air
distance = speed*time
2 = speed* .45
speed = 4.4 m/s horizontal
.008 v = .258 (4.4)
v = 141.9 m/s