唔識做物理功課??

In the first 15 m of fall, the string is still slack and has no tnesion in it.

The velocity v at the end of the 15 m fall is, using v^2 = u^2+2.a.s,
v = sqrt(2xgx15) m/s, where g is the acceleration due to gravity(=10 m/s2)
i.e. v = 17.32 m/s

The time needed t is, using v=u+.at
t = 17.32/g s = 17.32/10 s = 1.732 s

Since the total time spent is 4s, thus the time at which the string is under tension
= (4 - 1.732) s = 2.268 s

Using impulse = change in momentum
i.e. F.t = mv-mu
where F is the average retarding force, t is the time at which the force acts, m is the mass of the object, v and u are the final and initial velocities of the object.

F.(2.268) = 50x17.32 - 50x0
i.e. F = 50x17.32/2.268 N = 382 N

2007-03-16 13:58:32 補充：
The weight of the man has to be added, thus the total average tension in the string = (382+500) N = 882 N

s = (u+v)t/2
50 = (0+v)4/2
25 = v
v =25ms^(-1)

Tension (繩的平均拉力) = work done on KE
= (1/2)mv^2
= (1/2)(50)(25^2)
=15600N (3 sig. fig.)