A particle is projected vertically upwards with a speed of 40 m/s. At what times will it be 35m above its point of projection? \(\text{Take g} = 10 ms^{-2}\)
A. 1 sec and 7 sec B. 1 sec and 8 sec C. 2 sec and 5 sec D. 2 sec and 7 sec
Correct Answer: A
Explanation
\(s = ut + \frac{1}{2}at^{2}\) \(s = ut - \frac{1}{2}gt^{2}\) (Upward movement against gravity) \(35 = 40t - \frac{1}{2}10t^{2}\) \(35 = 40t - 5t^{2}\) \(5t^{2} - 40t + 35 = 0\) \(t^{2} - 8t + 7 = 0\) \((t - 1)(t - 7) = 0 \implies t = \text{1 sec and 7 sec}\)
