Question Detail

For red ball thrown down: s = ut + ½gt², with u = 30 m/s down. For green ball thrown up: it goes up and then down. Time for green to reach highest point: t_up = u/g = 30/9.8 ≈ 3.06 s. Time to fall from highest point back to cliff level is same 3.06 s, so total time to return to cliff level = 6.12 s. At that moment, red ball has been falling for 6.12 s and is far below. Green then continues falling from cliff level with zero initial velocity, taking additional time t such that the total height H is covered by red in less time. Actually, the question asks how long after red ball hits will green hit. If cliff height is H, red time t_red satisfies H = 30t_red + ½gt_red². Green time t_green = 2u/g + √(2H/g). The difference is 2u/g = 60/9.8 ≈ 6.12 s ≈ 6 s.