有關PHY MOTION的問題~~

Let h be the height at which the stone was dropped, and T be the time taken for the stone to reach the ground.

Since the stone travelled h/2 in the last second, hence it covered the rest of the distance h/2 in time (T-1)
Use equation: s = ut + (1/2)at^2
h/2 = (1/2)g(T-1)^2

For the whole journey, use the same equation again,
h = (1/2)gT^2

Dividing the two equations,
1/2 = [(T-1)/T]^2
i.e. (T-1)/T = 0.7071
T = 3.4142 s

Hence, h = (1/2).(10).(3.4142)^2 m = 58.28 m
[Take g = 10 m/s^2]