question_answer

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 \[m{{s}^{-1}},\] 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: \[(g=10\,m{{s}^{-2}})\]

A) 20 m

B) 30 m

C) 40 m    

D) 10 m

Correct Answer: C

Solution :

\[{{Y}_{cm}}\]from ground \[=\frac{0.03\times 100}{0.05}=60\,m\]

\[{{V}_{cm}}=\frac{0.02\times 100}{0.05}=40\,m/s\]

\[H=\frac{V_{cm}^{2}}{2g}=\frac{40\times 40}{20}=80\,m\]

Height above building\[=80-40=40\text{ }m.\]