Question: A body falling for 2 seconds covers a distance $S$ equal to that covered in next second. Taking \(...

Question

A body falling for 2 seconds covers a distance $S$ equal to that covered in next second. Taking $g = 10m/s^{2},S =$

Answer 1

If u is the initial velocity then distance covered by it in 2 sec

$S = ut + \frac{1}{2}at^{2} = u \times 2 + \frac{1}{2} \times 10 \times 4 = 2u + 20$ …(i)

Now distance covered by it in 3^rd sec

$S_{3^{rd}} = u + \frac{g}{2}(2 \times 3 - 1)10 = u + 25$ …(ii)

From (i) and (ii), $2u + 20 = u + 25 \Rightarrow u = 5$

$\therefore S = 2 \times 5 + 20 = 30\mspace{6mu} m$

Question: A body falling for 2 seconds covers a distance \(S\) equal to that covered in next second. Taking \(...