MCQ
A body falling freely under gravity passes two points $30\,m$ apart in $1\,s$ . From what point above the upper point it began to fall?......... $m$ (Take $g = 10\,m\,s^{-2}$ )
- ✓$31.25$
- B$16$
- C$8.6$
- D$4$
at $t$ second and lower point at $(t+1)$ s, then
$\mathrm{S}_{2}-\mathrm{S}_{1}=\frac{1}{2} \mathrm{g}(\mathrm{t}+1)^{2}-\frac{1}{2} \mathrm{gt}^{2}=\frac{1}{2} \mathrm{g}(2 \mathrm{t}+1)$
or $30 \mathrm{m}=\frac{1}{2} \times 10(2 \mathrm{t}+1)$
$\therefore \mathrm{t}=2.5 \mathrm{s}$
$S_{1}=\frac{1}{2} g t^{2}=\frac{1}{2} \times 10 \times(2.5)^{2}=31.25 \mathrm{m}$
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$Reason$ : If net external force is zero, then the linear momentum of the system changes