Question:
A stone is dropped from the top of a building. When it crosses a point $5 \mathrm{~m}$ below the top, another stone starts to fall from a point $25 \mathrm{~m}$ below the top, Both stones reach the bottom of building simultaneously. The height of the building is:
Correct Option: 1
Solution:
For particle (1) $20+h=10 t+\frac{1}{2} g t^{2} \quad \ldots . .(i)$
For particle (2) $h=\frac{1}{2} \mathrm{~g} t^{2} \quad \ldots$ (ii)
put equation (ii) in equation (i)
$20+\frac{1}{2} g t^{2}=10 t+\frac{1}{2} g t^{2}$
$t=2$ sec
Put in equation (ii)
$h=\frac{1}{2} g t^{2}$
$=\frac{1}{2} \times 10 \times 2^{2}$
$h=20 \mathrm{~m}$
the height of the building $=25+20=45 \mathrm{~m}$