An 82 kg man lowers himself to the ground from a height of 6.6 m by holding onto a rope that runs over a frictionless pulley to a 66 kg sandbag. With what speed does the man hit the ground if he started from rest?

Answer 1

Force up on 66 kg mass = T
Force down on 66 kg mass = 66(9.8 )
net force up on 66 kg = T-66(9.8)
F = m a
T -66(9.8) = 66 a
a is down

Force down on 82 kg mass = 82(9.8)
Force up on 82 kg mass = T
net force down on 82 kg = 82(9.8) -T
F = m a
82(9.8) - T = 82 a
combine by addition
(82-66)(9.8) = (82+66) a
solve for a which is down
then
6.6 = (1/2) a t^2
solve for t
then v = a t