A helicopter 7.60 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.

A)Just as it hits the ground, find the velocity of the package relative to the helicopter.
vph=?m/s

B)Just as it hits the ground, find the velocity of the helicopter relative to the package
Vph=?m/s

b. odd question. Is the helicopter desending at a constant rate of 3.9m/s? Helicopters do not ordinarily Fall under the influence of gravity.

a. Vf^2=Vi^2+2gh
Vf^2=3.9^2 + 2*9.8*7.6

relative to helicopter? subtract helicopter speed.

for A is it 164.17

b. would be a upward direction

i didn't put 7.60m and 3.90m/s in the question problem

well i square rooted 164.17 and gave me 12.8128 then i put - 12.8 and it gave me a wrong anwser for my hw problem

To solve this problem, we need to consider the motion of both the helicopter and the package separately.

A) Just as the package hits the ground, the velocity of the package relative to the helicopter can be found using the equation:

vf = vi + at

where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. In this case, since the package falls freely and no external forces act on it, its acceleration is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2.

Since the package starts from rest (relative to the helicopter), its initial velocity (vi) is zero. The time it takes for the package to hit the ground can be found using the equation:

d = vit + 0.5at^2

where d is the distance, vi is the initial velocity, a is the acceleration, and t is the time. In this case, the distance (d) is equal to the initial height of the package above the ground, which is 7.60 m.

We can find the time (t) by rearranging the equation:

t = sqrt(2d/a)

Plugging in the values, we get:

t = sqrt(2 * 7.6 / 9.8) = 1.08 s

Now, we can calculate the final velocity (vf) using the equation:

vf = vi + at

Since vi is zero and a is -9.8 m/s^2 (negative because the acceleration is opposite to the positive direction), we have:

vf = 0 + (-9.8) * 1.08 = -10.58 m/s

Therefore, the velocity of the package relative to the helicopter just as it hits the ground is approximately -10.58 m/s (downward).

B) To find the velocity of the helicopter relative to the package just as it hits the ground, we can use the fact that the velocity of an object (helicopter) relative to another object (package) is equal in magnitude but opposite in direction to the velocity of the second object (package) relative to the first object (helicopter).

So, the velocity of the helicopter relative to the package is equal in magnitude to the velocity of the package relative to the helicopter, but in the opposite direction. Therefore, the velocity of the helicopter relative to the package is approximately 10.58 m/s (upward).