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?

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

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

Where:
h = height of the object (90m)
u = initial velocity of the object (0 m/s since the package is dropped, not thrown)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time

By rearranging the equation, we can solve for time:

t = sqrt((2h) / g)

Substituting the values:

t = sqrt((2 * 90) / 9.8)
t ≈ sqrt(18.36)
t ≈ 4.28 seconds

Therefore, it takes approximately 4.28 seconds for the package to hit the ground.

To find the speed of the package just before impact, we can use the formula:

v = u + gt

Where:
v = final velocity (speed before impact)
u = initial velocity (0 m/s)
g = acceleration due to gravity (9.8 m/s^2)
t = time (4.28 seconds)

Substituting the values:

v = 0 + (9.8 * 4.28)
v ≈ 41.94 m/s

Therefore, the package is moving at approximately 41.94 m/s just before it hits the ground.