A helicopter is ascending (moving upwards) at a constant velocity of 8 m/s. At a height of 90m, a passenger opens a window and drops a package to the ground below. How long does it take the package to hit the ground and how fast is it moving just before impact?

d = Vo*t + 0.5gt^2 = 90m,

0 + 0.5*9.8*t^2 = 90,
4.9t^2 = 90,
t^2 = 18.36,
t = 4.29s.

Vf = Vo + gt,
Vf = 0 + 9.8*4.29 = 42m/s.

To find out how long it takes for the package to hit the ground, we can use the equation of motion:

h = ut + (1/2)gt^2

Where:
h = height (90m in this case)
u = initial velocity (since the package is dropped, its initial velocity is 0)
g = acceleration due to gravity (-9.8 m/s^2)
t = time

At the point of impact, the height is 0, so we can rewrite the equation as:

0 = 0 + (1/2)(-9.8)t^2

Simplifying the equation:

0 = -4.9t^2

Dividing both sides by -4.9:

0 = t^2

Since the square of a number is zero only when the number itself is zero, we know that t = 0.

Therefore, it takes 0 seconds for the package to hit the ground.

Now, let's calculate the speed of the package just before impact.

The speed of an object in free fall can be calculated using the equation:

v = u + gt

Where:
v = final velocity
u = initial velocity (0 in this case)
g = acceleration due to gravity (-9.8 m/s^2)
t = time (0 seconds in this case)

Substituting the values into the equation:

v = 0 + (-9.8)(0)
v = 0

Therefore, the package has a speed of 0 m/s just before impact.

To find the time it takes for the package to hit the ground, we can use the equation of motion:

s = ut + (1/2)at^2

where:
s = displacement (height of the helicopter = 90m)
u = initial velocity (velocity of the helicopter = 8 m/s)
a = acceleration (acceleration due to gravity = -9.8 m/s^2, negative because it is acting in the opposite direction)
t = time

Since the helicopter is ascending, the displacement is positive. Therefore, the equation becomes:

90 = 8t + (1/2)(-9.8)t^2

Now, we need to solve this quadratic equation to find the time it takes for the package to hit the ground. Since the coefficient of t^2 is negative, we'll have two solutions for t. We can disregard the negative value since time cannot be negative in this context.

Simplifying the equation, we get:

4.9t^2 - 8t - 90 = 0

To solve for t, we can use the quadratic formula:

t = (-b + √(b^2 - 4ac)) / (2a)

where:
a = 4.9
b = -8
c = -90

Substituting these values and solving the equation, we get:

t ≈ 4.93 seconds

Therefore, it takes approximately 4.93 seconds for the package to hit the ground.

To find the final velocity of the package just before impact, we can use the equation:

v = u + at

where:
v = final velocity
u = initial velocity (0 m/s since the package was dropped from rest)
a = acceleration due to gravity (-9.8 m/s^2)

Substituting these values, we get:

v = 0 + (-9.8)(4.93)

Simplifying, we find:

v ≈ -48.15 m/s

The negative sign indicates that the velocity is directed downwards. Therefore, the package is moving with a speed of approximately 48.15 m/s just before impact.