A helicopter 8.70 above the ground and descending at 2.60 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.
Vph=

Just as it hits the ground, find the velocity of the helicopter relative to the package.
Vhp=

Hello i just worked on this problem, Vf^2=Vo^2+2gh

Vf- velocity final
Vo-Initial velocity
g-gravity
h-height

Vf^2=(2.60^2) + 2*(9.8)(8.70)
Vf^2=177.28
square root it... 13.314m/s
Subtract from heli speed, 13.314-2.60= -10.714
negative because up is positive and down well is negative lol

hope it works, that's what i did for my gf homework and she got it right!

Still working on part B sorry :/

To solve this problem, we can use the equations of motion for objects in free fall.

First, let's calculate the time it takes for the package to reach the ground. We can use the equation:

h = (1/2)gt^2

where h is the initial vertical distance above the ground, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

Given that the initial height of the package is 8.70 m, we can solve for t:

8.70 = (1/2)(9.8)t^2
t^2 = (2 * 8.70) / 9.8
t^2 = 1.7755
t = √1.7755
t = 1.33 seconds (approximately)

Now, let's find the velocity of the package just as it hits the ground. We can use the equation:

v = gt

where v is the velocity, g is the acceleration due to gravity, and t is the time.

Plugging in the values:

v = (9.8)(1.33)
v = 13.03 m/s (approximately)

Therefore, the velocity of the package just as it hits the ground, relative to the helicopter, is 13.03 m/s downward.

Now, let's find the velocity of the helicopter relative to the package just as it hits the ground. Since the package and the helicopter descend together, their velocities will be the same.

Therefore, the velocity of the helicopter relative to the package, just as the package hits the ground, is also 13.03 m/s downward.