Ok, lets call up positive and down negative.

The package has an initial velocity of +5.2m/s and an initial height of +125m (assuming the ground is 0m). We know the acceleration will be that of gravity (-9.81m/s^2). We can therefore use the following equation to find the velocity when it hits the ground:

vf^2 = vi^2 + 2ad

vf^2 = (5.2)^2 + 2(-9.81)(0 – 125)

vf = -49.79m/s

(the negative sign just indicates the downward direction)

Now we find the time using the following equation:

vf = vi + at

-49.79 = 5.2 + (-9.81)t

t = 5.61s

just in 9th grade but im pretty sure that the part about the rate of the helicopters ascending is irrelevant to finding out how fast the package will drop to the ground, you just need to be told at what height the box was dropped. Remember also the time it takes for the package to drop to the ground isnt affected by its weight or w.e. Also in the answer im pretty sure you would see the speed at which the box falls increases more and more the longer it is freefall (though I think it maxes out at some huge number/not sure though/probably doesnt apply to this problem), as the box continues to accelerate from the pull of gravity + its current acceleration. Hope this helps somewhat, or just to give you a better understanding of what the problem will figure out to require, you might know all of this alreaedy though.

s=vt+0.5at^2

125 = -5.20t +0.5gt^2

125 = -5.2t + (0.5*9.8)t^2

125=-5.2t + 4.9t^2

0=-125 – 5.2t + 4.9t^2

t = (-b+/-sqrt(b^2-4ac) )/2a

t = (5.2+/-sqrt(5.2^2 – 4(4.9)(-125)) )/(2(4.9))