A helicopter 9.00 above the ground and descending at 3.90 drops a package from rest (relative to the helicopter). We have chosen the positive positive direction to be upward. The package falls freely.
Just as it hits the ground, find the velocity of the package relative to the helicopter.
Just as it hits the ground, find the velocity of the helicopter relative to the package.

Question

A helicopter 9.00 above the ground and descending at 3.90 drops a package from rest (relative to the helicopter). We have chosen the positive positive direction to be upward. The package falls freely.
Just as it hits the ground, find the velocity of the package relative to the helicopter.
Just as it hits the ground, find the velocity of the helicopter relative to the package.

vincent-lyon.fr · Answer

9.00 what?
3.90 what?

anonymous · Answer

Supossing it's 9 meters above the ground and it's descending at a constant speed of 3.9 m/s .
$$v^2 =v_0^2 - 2a(y)$$
$$v =\sqrt{(3.9)^2 +2(9.8)(9.0)} = 13.84 m/s$$ with respect to the ground.
Since you're taking upward as positive, the velocity is -13.84 m/s .
With respect to the helicopter 
$$V_A = V_1 - V_2 $$
$$V_A = -13.84 - (-3.9) = 9.94 m/s$$
when it hits the ground.