Question

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?

Answer 1

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 $x+100\text{ m} = 286+100\text{ m} = 386\text{ m}$

Answer 2

Answer:

100

Explanation:

this is because we know the plane is flying horizontally at a speed of 50 m/s. and we know that after 2.0 seconds, it drops a package. in 2.0 seconds, the speed is 50m/s. We then multiply these values of 50(2) and we get 100 m.