a
(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), \(2u + 20 = u + 25 \Rightarrow u = 5\)

\(\therefore S = 2 \times 5 + 20 = 30\;m\)