Ahelicopter is ascending vertically with a speed of 7.00 m/s. at a height of 155 m above the earth, a package is dropped from a window. how much time does it take for the package to reach the ground? [hint: the package's initial speed equals the helicopter's.]

Given:
u = 7.00m/s, the initial vertical velocity of the package.

Assume that g = 9.8 m/s².
Neglect air resistance.
All quantities are measured as positive upwards.

The height from which the package is dropped is 155 m above ground.
Let t = the time for the package to hit the ground.
Therefore
(7.00 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (- 155 m)
4.9t² - 7t - 155 = 0
or
t² - 1.4286t - 31.6327  0
Solve with the quadratic formula.
t = (1/2)*[1.4286 +/- √(2.0409 + 126.5308)]
  = 6.384 s or - 4.955 s
Reject negative time.

6.384 s