A piece of wood of mass $0.03\ \text{kg}$ is dropped from the top of a $100\ \text{m}$ height building. At the same time, a bullet of mass $0.02\ \text{kg}$ is fired vertically upward, with a velocity $100\ \text{m s}^{-1}$, 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: $(g = 10\ \text{m s}^{-2})$
Answer: (A) $40\ \text{m}$
Relative to each other the two move at $100\ \text{m/s}$ (gravity affects both equally), so they meet after $t = \dfrac{100}{100} = 1\ \text{s}$.
At $t = 1\ \text{s}$: wood speed $10\ \text{m/s}$ down, having fallen $5\ \text{m}$ (so it is $95\ \text{m}$ above ground); bullet speed $100 - 10 = 90\ \text{m/s}$ up.
Momentum conservation (up positive):
$$0.02(90) - 0.03(10) = 0.05\,v \;\Rightarrow\; v = 30\ \text{m/s upward}$$
Further rise $= \dfrac{30^2}{2(10)} = 45\ \text{m}$, so the highest point is $95 + 45 = 140\ \text{m}$, which is $40\ \text{m}$ above the top of the building.
Solution by Sreeraj P, M.Sc Physics