✔ 最佳答案
Given: mass of person = 70 kg, falling height = 62 m
Use equation of motion: v^2 = u^2 + 2a.s
with u = 0 m/s, a = g(=9.81 m/s^2), s = 62 m v =?
hence, v^2 = 2 x 9.81 x 62 (m/s)^2
i.e. v = 34.88 m/s = 126 km/h
Speed before striking the water surface = 126 km/h
Use equation of motion: s = ut + (1/2)at^2
with s = 62 m, u = 0 m/s, a = g(=9.81 m/s^2), t =?
hence, 62 = (1/2).(9.81)t^2
t = 3.56 s
The time of fall = 3.56 s
You need to make an assumption on the impact time on the water surface, just assume it to be 1 s (in order of magnitude).
Use equation: impulse (衝量) = change of momentum (動量改變)
(F - 70g) x 1 = 70 x 34.88
where F is the impact force (衝擊力)
solve for F gives F = 3128 N = 3128/9.81 kgf = 318.9 kgf = 318.9 x 2.2 lbf
= 702 lbf