A helicopter is flying horizontally at 7.5 m/s and an altitude of 18 m when a package of emergency medical supplies is ejected horizontally backward with a speed of 13 m/s relative to the helicopter. Ignoring air resistance, what is the horizontal distance between the package and the helicopter when the package hits the ground?

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This almost exact problem was answered.

To find the horizontal distance between the package and the helicopter when the package hits the ground, we can use the formula:

\[d = v_x \cdot t\]

where:
- \(d\) is the horizontal distance
- \(v_x\) is the horizontal velocity
- \(t\) is the time it takes for the package to hit the ground

First, let's find the time it takes for the package to hit the ground. Since the package is ejected horizontally with a speed of 13 m/s, its vertical velocity is 0 m/s (as there is no initial vertical velocity). We can use the formula for vertical displacement:

\[y = v_{iy} \cdot t + \frac{1}{2} \cdot g \cdot t^2\]

Since the package is falling downward, the vertical displacement is -18 m, the initial vertical velocity is 0 m/s, and the acceleration due to gravity is approximately 9.8 m/s². Using these values, the equation becomes:

\[-18 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2\]

Simplifying the equation and solving for \(t\), we get:

\[9.8t^2 = 18\]
\[t^2 = \frac{18}{9.8}\]
\[t^2 \approx 1.8367\]
\[t \approx 1.3568 \text{ seconds}\]

Now, we can find the horizontal distance by multiplying the horizontal velocity by the time:

\[d = 7.5 \cdot 1.3568\]
\[d \approx 10.2056 \text{ meters}\]

Therefore, the horizontal distance between the package and the helicopter when the package hits the ground is approximately 10.21 meters.