Physics Force&Motion 物力 力學

Let 2h be the height of the tower, and T be the total time taken to fall from the top of the tower to ground.
Thus, time taken to fall from the top to the mid-point of the tower = (T-0.4)

Consider the whole journey, use equation of motion: s = ut + (1/2)at^2
with u = 0 m/s, a = g(=10 m/s2), t = T, s = 2h
hence, 2h = (10/2)T^2
i.e. 2h = 5T^2
or h = (5/2)T^2 --------------------- (1)

Consider the journey from the top to the mid-point, use equation of motion: s = ut + (1/2)at^2
with u = 0 m/s, a = g(= 10 m/s2), t = (T-0.4), s = h
hence, h = (10/2)(T-0.4)^2
i.e. h = 5T^2 - 4T + 0.8 --------------------- (2)

Equating (1) and (2): 5T^2 = 2( 5T^2 - 4T + 0.8 )
i.e. 5T^2 -8T + 1.6 = 0
solve for T gives T = 1.3657 or 0.2343 (rejected, because shorter than 0.4s)

Therefore, the time of fall is 1.37 s