A challenge!

Something wrong in the answer provided by zip852hk for the part of finding the location.

For the time:
The only force acting on the crate is the gravitational pull. (Air resistance is neglected here) So,
s = ut + 1/2 at²
8500 = (0)t + 1/2 (g) t²【When released, the crate is VERTICALLY at rest】
t² = 17000/g
t = √(17000/g)
◆If you take g = 10 ms-2,
　t = √(17000/10) = 41.2 sec (3 sig. fig.)
◆If you take g = 9.8 ms-2,
　t = √(17000/9.8) = 41.6 sec (3 sig. fig.)

For the landing place:
The only force acting on the crate is the wind (250N). (Air resistance is also neglected here)
BUT NOTE THAT since the plane is travelling at 120 ms-1 when the crate is released, the crate has an initial horizontal speed of 120 ms-1 towards east.
From Newton's Second Law,
F = ma
250 = 150 a
a = 5/3 ms-2

s = ut + 1/2 at²
s = (120) √(17000/g) + 1/2 (-5/3) √(17000/g)²
【Since the wind is blowing WEST, the acceleration is opposite in direction with motion, a negative sign is added. t is the value taken from the above part】
s = (120) √(17000/g) - 42500/(3g)
◆If you take g = 10 ms-2,
　s = (120) √(17000/10) - 42500/(3 x 10) = 3530 m (3 sig. fig.)【East of release location】
◆If you take g = 9.8 ms-2,
　s = (120) √(17000/9.8) - 42500/(3 x 9.8) = 3550 m (3 sig. fig.)【East of release location】


Hope it helps! ^^
Wish you a Happy 2007!

very easy
first, consider vertically
s=ut+1/2at^2(u=0)
8500=1/2x10xt^2
t=41.2s
so u know when

and now what is the position?
consider the horizontal force
F=ma
250=150a
a=1.67
s=ut+1/2at^2(u=0again)
as it is released, not pushed
s=1/2x1.67x41.2^2
s=1420m
therefor the position is 1420m east from original point

i hope i didnt think wrong