A helicopter is ascending vertically with a speed of 5.20 m/s. At a height of 125 m above the Earth, a package is dropped from a window. How much time does it takes the package to hit the ground? [Hint: The package's initial speed equals the helicopter's]

h=v²/2g=5.2²/2• 9.8=1.38 m.

H=125+1.38=126.38 m
t=sqrt(2•H/g) = sqrt(2•126.38/9.8)=5.08 s.

To find the time it takes for the package to hit the ground, we can use the equation of motion for vertical motion under constant acceleration.

The initial velocity of the package is the same as the helicopter's velocity, which is 5.20 m/s in the upward direction. Since the package is dropped, it will accelerate downward due to gravity. The acceleration due to gravity near the Earth's surface is approximately 9.8 m/s² and acts in the downward direction.

We can use the equation:

s = ut + (1/2)at²

where:
s = vertical displacement
u = initial velocity
t = time
a = acceleration

In this case, s is the height above the ground (125 m), u is the initial upward velocity of the helicopter (5.20 m/s), a is the acceleration due to gravity (-9.8 m/s²), and we want to solve for t.

Plugging in the values, we have:

125 = (5.20)t + (1/2)(-9.8)t²

Simplifying the equation, we get:

125 = 5.20t - 4.9t²

Rearranging the equation:

4.9t² - 5.20t + 125 = 0

This is a quadratic equation. We can solve it by using the quadratic formula:

t = (-b ± √(b² - 4ac)) / 2a

For our equation:

a = 4.9
b = -5.20
c = 125

Plugging these values into the quadratic formula, we can find the time.

t = (-(-5.20) ± √((-5.20)² - 4(4.9)(125))) / (2 * 4.9)

Calculating further:

t = (5.20 ± √(27.04 + 2450)) / 9.8
t = (5.20 ± √(2477.04)) / 9.8
t = (5.20 ± 49.77) / 9.8

We have two possible options for t:

t1 = (5.20 + 49.77) / 9.8
t2 = (5.20 - 49.77) / 9.8

Evaluating each option:

t1 = 54.97 / 9.8 ≈ 5.61 s
t2 = -44.57 / 9.8 ≈ -4.55 s

Since time cannot be negative, we discard the negative solution. Therefore, the package takes approximately 5.61 seconds to hit the ground.