A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 m/s, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is :(g=10 m/s²).

Time taken for the particles to collide, t = d/Vrel = 100/100 = 1 sec
Speed of wood just before collision = gt = 10 m/s
and speed of bullet just before collision v - gt = 100 m/s - 10 m/s = 90 m/s
Now, conservation of linear momentum just before and after the collision -
-(0.02)(1v) + (0.02)(9v) = (0.05)v ⇒ 150 = 5v ⇒ v = 30 m/s
Max. height reached by body h = v²/2g
Before : 0.03 kg ↓ 10 m/s
0.02 kg ↑ 90 m/s
After : v 0.05 kg
h = 30 × 30 / 2 × 10 = 40m.