A piece of wood of mass 0.03   k g is dropped from the top of a 100   m height building. At the…

1. $40 m$
2. $20 m$
3. $10 m$
4. $30 m$

Solution

Using relative velocity



$a_{\mathrm{r}\mathrm{e}\mathrm{l}}=0$

$v_{\mathrm{r}\mathrm{e}\mathrm{l}}=100$

Time $t$ for collision,

$100=v_{\mathrm{r}\mathrm{e}\mathrm{l}}\times t$

$t=\frac{100}{100}=1\mathrm{s}\mathrm{e}\mathrm{c}$

$v_{\mathrm{b}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{e}\mathrm{t}}=100-1\times 10=90m/s$

$v_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{l}\mathrm{e}}=10\times 1=10m/s$

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



$P_i=P_f$

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

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

So from top of building

$45-5=40m$

Asked in: JEE Main 2019 (10 Jan Shift 1)

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