Question
A body falling for $2 \,seconds$ covers a distance $S$ equal to that covered in next second. Taking $g = 10\,m/{s^2},\,S =..........m$
✓
Answer
(a) If $u$ is the initial velocity then distance covered by it in $2$ sec
$S = ut + \frac{1}{2}a{t^2} = u \times 2 + \frac{1}{2} \times 10 \times 4 = 2u + 20$ ... (i)
Now distance covered by it in $3^{rd}$ sec
${S_{{3^{rd}}}} = u + \frac{g}{2}\left( {2 \times 3 - 1} \right)10 = u + 25$ …(ii)
From(i) and (ii), $2u + 20 = u + 25 \Rightarrow u = 5$
$\therefore S = 2 \times 5 + 20 = 30\;m$
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