Ahelicopter is ascending vertically with a speed of 5.52 m/s . at a height of 125 m above the earth, a package is dropped from the helicopter. how much time does it take for the package to reach the ground?

This is a perfect application for the general formula for the height of an object in gravity at any time:

     Height =        (original height)
                       + (original velocity x time)
                       - (1/2 x gravity x time²)

         H  =  H₀ + v₀T - 1/2 G T²

In this helicopter problem:

H₀ = 125 m
v₀ = 5.52 m/s
G = 9.8 m/s²

and we want to find 'T' when the package hits the ground.
That's the time when H=0 . 

           H₀ + v₀T - 1/2 G T² = 0

         125 + 5.52T - 4.9T² = 0

Using the quadratic formula:
    
         T = -5.52 ± √[5.52² + (4 x 4.9 x 125) ]  all over (-9.8)  

             = -5.52 ± √2480.47        all over (-9.8)

             = 0.563  ±  5.082

         T  =  -4.52
         T  =   5.65

In a real-world situation, we ignore the negative solution.  

The package hits the ground  5.65 seconds  after being released.

I hope there was nothing fragile inside.

    5.5 n======    assuming that      we have      m = 0.145 kg    a = 38 m/s²therefore    5.51 n, but two significant figures.final answer: 5.5 n