A passenger in a helicopter traveling upwards at 17 m/s accidentally drops a package out the window. If it takes 9 seconds to reach the ground, how high to the nearest meter was the helicopter when the package was dropped?To the nearest meter what was the maximum height of the package above the ground in the previous problem?

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To find the height of the helicopter when the package was dropped, we can use the formula for vertical motion:

h = h0 + v0t + 0.5gt^2

Where:
h = final height of the helicopter
h0 = initial height of the helicopter
v0 = initial velocity of the helicopter (in this case, 0 m/s because it was already at its final velocity when the package was dropped)
t = time taken (9 seconds)
g = acceleration due to gravity (approximately 9.8 m/s^2)

Now, let's substitute the known values into the equation:

h = h0 + v0t + 0.5gt^2
h = h0 + 0 + 0.5 * 9.8 * 9^2
h = h0 + 0 + 0.5 * 9.8 * 81
h = h0 + 0 + 397.35

Since the package took 9 seconds to reach the ground, the initial height of the helicopter can be calculated by subtracting the distance traveled in 9 seconds (assuming constant velocity) from the final height.

Let's calculate the distance traveled in 9 seconds:

d = v * t
d = 17 m/s * 9 s
d = 153 m

Now, we can find the initial height:

h0 = h - d
h0 = 397.35 - 153
h0 ≈ 244.35 m

So, the helicopter was approximately 244 meters high when the package was dropped.

To find the maximum height of the package above the ground, we need to consider the time it takes for the package to reach the ground. We already know that it took 9 seconds to reach the ground, so we can use the formula:

h = h0 + v0t + 0.5gt^2

Since the package starts from rest (v0 = 0), the formula simplifies to:

h = h0 + 0.5gt^2

Substituting the known values:

h = 0 + 0.5 * 9.8 * 9^2
h = 0.5 * 9.8 * 81
h ≈ 400.05 m

Therefore, the maximum height of the package above the ground in the previous problem was approximately 400 meters.