有什麼方法計到?????????

其實要解呢條問題唔系要知28點來，系要知呢條問題是供的資料作分析：
每爬3米就下跌1米,每爬1米要1秒,每下跌1米也要1秒
即是爬1米或2米唔會下跌，只有最接近30米時先可以只爬1米或2米，咁即系28米或29米時唔會下跌
即系頭28米或29米時，每爬２米要４秒，由於每次系爬2米，所以冇可能會下跌到２９米，所以一定系２８米時再用２秒爬上３０米
所以系［（２８／２）Ｘ４］＋２

Q1.For 2 seconds,
Distance = 1x2 = 2m
Speed = 2/2 = 1m/s

For 4 seconds,
Distance = 1x3 - 1 = 2m
Speed = 2/4 = 0.5m/s

For 28 meters,
Time = 28/0.5 = 56s

56/4 = 14
the monkey drops at t = 56s

For the last 2 meters,
Time = 2/1 = 2s

Total time cost = 2 + 56 = 58s

28 is the first 28m and the last drop of the monkey is ended.

Q2.Some may think the time needed is 30/0.5 = 60s, but the monkey arrives 30m, it is no need to count its drop after 2s.