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vredina [299]
3 years ago
10

A plane flying horizontally at a speed of 50 m/s and at an elevation of 160 m drops a package, and 2.0 s later it drops a second

package. How far apart will the two packages land on the ground if air resistance is negligible?
Physics
1 answer:
Anarel [89]3 years ago
7 0

Answer:

386 m

Explanation:

Let's call the horizontal distance between the point of launch and the point of landing of the first package x\text{ m}.

2 seconds after landing at x, the plane has travelled a horizontal distance of 2\text{ s}\times 50 \text{ m/s} = 100\text{ m}.

At this new point, the second package is launched. Because it is launched under the same conditions as the first, its horizontal distance from its point of launch is also x\text { m}.

To determine x, we determine the time of fall of the package. Being a vertical motion, the drop has an initial velocity, u, of 0 m/s with acceleration, a = g = 9.8 \text{ m/s}{}^2, and a distance, s, of 160 m. We use the equation of motion:

s = ut+\frac{1}{2}at^2

160 = 0\times t + \frac{1}{2}9.8\times t^2

t = \sqrt{\dfrac{160}{4.9}}\text { s}=\dfrac{40}{7}\text { s}

In this time, the package, with a horizontal velocity of 50 m/s, travels the horizontal distance, x, of

x = 50 \text{ m/s}\times \dfrac{40}{7}\text { s} = \dfrac{2000}{7}\text { m} = 286 \text{ m}

The distance apart between both packages is x+100\text{ m} = 286+100\text{ m} = 386\text{ m}

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