1 sec and 7 sec
1 sec and 8 sec
2 sec and 5 sec
2 sec and 7 sec
Correct answer is A
\(s = ut + \frac{1}{2}at^{2}\)
\(s = ut - \frac{1}{2}gt^{2}\) (Upward movement against gravity)
\(35 = 40t - \frac{1}{2}10t^{2}\)
\(35 = 40t - 5t^{2}\)
\(5t^{2} - 40t + 35 = 0\)
\(t^{2} - 8t + 7 = 0\)
\((t - 1)(t - 7) = 0 \implies t = \text{1 sec and 7 sec}\)