An airplane traveling 1011 m above the ocean at 145 km/h is going to drop a box of supplies to shipwrecked victims below.

(a) How many seconds before the plane is directly overhead should the box be dropped?
1 s

(b) What is the horizontal distance between the plane and the victims when the box is dropped?

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To determine the horizontal distance between the plane and the victims when the box is dropped, we can use the formula for horizontal distance:

Distance = Speed x Time

In this case, the speed of the plane is given as 145 km/h. However, we need to convert this to meters per second to match the units of the given vertical distance (1011 m) above the ocean.

To convert km/h to m/s, we can use the conversion factor: 1 km/h = 0.2778 m/s

So, the speed of the plane in meters per second is:
145 km/h * 0.2778 m/s = 40.278 m/s (rounded to 3 decimal places)

Now, we can calculate the time it takes for the box to drop by using the distance formula in the vertical direction:

Distance = (1/2) * g * t^2

In this situation, the given vertical distance is 1011 m and we want to find the time, t. The acceleration due to gravity, g, is approximately 9.8 m/s^2.

Rearranging the formula, we have:

t^2 = (2 * Distance) / g

Substituting the given values:

t^2 = (2 * 1011 m) / 9.8 m/s^2

Simplifying:

t^2 = 206.327551 m

Taking the square root of both sides, we find:

t ≈ 14.358 s (rounded to 3 decimal places)

Finally, to determine the horizontal distance between the plane and the victims when the box is dropped, we can multiply the horizontal speed of the plane by the time it took for the box to drop:

Horizontal Distance = Speed x Time

Horizontal Distance = 40.278 m/s * 14.358 s

Horizontal Distance ≈ 578.315 m (rounded to 3 decimal places)

Therefore, the horizontal distance between the plane and the victims when the box is dropped is approximately 578.315 meters.