Question 25 of 130

A piece of wood of mass $0.03kg$ is dropped from the top of a $100m$ height building. At the same time, a bullet of mass $0.02kg$ is fired vertically upward, with a velocity $100m{s}^{-1},$ 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: $\left(g=10m{s}^{-2}\right)$

A$40m$B$20m$C$10m$D$30m$

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Answer: (A) $40m$

Using relative velocity



${a}_{rel}=0$

${v}_{rel}=100$

Time $t$ for collision,

$100={v}_{rel}\times t$

$t=\dfrac{100}{100}=1 sec$

${v}_{bullet}=100-1\times 10=90m/s$

${v}_{particle}=10\times 1=10m/s$

$S=100\times 1-\dfrac{1}{2}\times 10\times 1=95 m$



${P}_{i}={P}_{f}$

$\Rightarrow 90\times 0.02-10\times 0.03=0.05V$

$V=30m/s\Rightarrow h=\dfrac{{v}^{2}}{2g}=\dfrac{900}{20}=45 m$

So from top of building

$45-5=40 m$