Free-Falling Physics help!!!

1. A helicopter is ascending vertically with a speed of 6.50 m/s. At a height of 185 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.

2. A stone is thrown vertically upward with a speed of 10.0 m/s from the edge of a cliff 86.0 m high.
A. How much later does it reach the bottom of the cliff?
B. What is its speed just before hitting?
C. What total distance did it travel?

I am stuck on these two problems so if you can please please please please help!!! THank you!!

Height package continues to rise = (v^2/2g), = (6.5^2/19.6), = 2.156m. Then, it drops.
Drop height = (185 + 2.156) = 187.156 metres.
Time to hit ground = sqrt. (2h/g), = 6.18 secs.

Similar problem. Height above cliff it rises to = (v^2/2g), = 5.102m.
Time rising from launch = (v/g) = 10/9.8, = 1.02 secs.
Total fall height = (86 + 5.102) = 91.102m.
Time from max. height to ground = sqrt. (2h/g), = 4.312 secs.
a) Time in air = 4.312 + 1.02, = 5.332 secs.
b) Velocity at ground = (gt), = 9.8 x 4.312, = 42.258 m/sec.
It travelled 185 + (2 x 5.102) = 195.204 metres.

1) Let: downward direction be -ve and the upward direction be +ve.

Given that a= (v-vi)/t; time then: t= (v-vi)/a, where downward is -ve acceleration in the the same direction is also -ve.

t=(v-vi)/-a => [(0- 6.5)m/s]/- (9.81)m/s^2 = 0.663s