# 3.5 Addition of velocities  (Page 4/12)

 Page 4 / 12

## Calculating relative velocity: an airline passenger drops a coin

An airline passenger drops a coin while the plane is moving at 260 m/s. What is the velocity of the coin when it strikes the floor 1.50 m below its point of release: (a) Measured relative to the plane? (b) Measured relative to the Earth?

Strategy

Both problems can be solved with the techniques for falling objects and projectiles. In part (a), the initial velocity of the coin is zero relative to the plane, so the motion is that of a falling object (one-dimensional). In part (b), the initial velocity is 260 m/s horizontal relative to the Earth and gravity is vertical, so this motion is a projectile motion. In both parts, it is best to use a coordinate system with vertical and horizontal axes.

Solution for (a)

Using the given information, we note that the initial velocity and position are zero, and the final position is 1.50 m. The final velocity can be found using the equation:

${{v}_{y}}^{2}={{v}_{0y}}^{2}-2g\left(y-{y}_{0}\right)\text{.}$

Substituting known values into the equation, we get

${{v}_{y}}^{2}={0}^{2}-2\left(9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)\left(-1\text{.}\text{50}\phantom{\rule{0.25em}{0ex}}\text{m}-0 m\right)=\text{29}\text{.}4\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}{\text{/s}}^{2}$

yielding

${v}_{y}=-5\text{.}\text{42 m/s.}$

We know that the square root of 29.4 has two roots: 5.42 and -5.42. We choose the negative root because we know that the velocity is directed downwards, and we have defined the positive direction to be upwards. There is no initial horizontal velocity relative to the plane and no horizontal acceleration, and so the motion is straight down relative to the plane.

Solution for (b)

Because the initial vertical velocity is zero relative to the ground and vertical motion is independent of horizontal motion, the final vertical velocity for the coin relative to the ground is ${v}_{y}=-5.42\phantom{\rule{0.25em}{0ex}}\text{m/s}$ , the same as found in part (a). In contrast to part (a), there now is a horizontal component of the velocity. However, since there is no horizontal acceleration, the initial and final horizontal velocities are the same and . The x - and y -components of velocity can be combined to find the magnitude of the final velocity:

$v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}\text{.}$

Thus,

$v=\sqrt{\left(\text{260 m/s}{\right)}^{2}+\left(-5\text{.}\text{42 m/s}{\right)}^{2}}$

yielding

$v=\text{260}\text{.}\text{06 m/s.}$

The direction is given by:

$\theta ={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right)={\text{tan}}^{-1}\left(-5\text{.}\text{42}/\text{260}\right)$

so that

$\theta ={\text{tan}}^{-1}\left(-0\text{.}\text{0208}\right)=-1\text{.}\text{19º}\text{.}$

Discussion

In part (a), the final velocity relative to the plane is the same as it would be if the coin were dropped from rest on the Earth and fell 1.50 m. This result fits our experience; objects in a plane fall the same way when the plane is flying horizontally as when it is at rest on the ground. This result is also true in moving cars. In part (b), an observer on the ground sees a much different motion for the coin. The plane is moving so fast horizontally to begin with that its final velocity is barely greater than the initial velocity. Once again, we see that in two dimensions, vectors do not add like ordinary numbers—the final velocity v in part (b) is not ; rather, it is . The velocity’s magnitude had to be calculated to five digits to see any difference from that of the airplane. The motions as seen by different observers (one in the plane and one on the ground) in this example are analogous to those discussed for the binoculars dropped from the mast of a moving ship, except that the velocity of the plane is much larger, so that the two observers see very different paths. (See [link] .) In addition, both observers see the coin fall 1.50 m vertically, but the one on the ground also sees it move forward 144 m (this calculation is left for the reader). Thus, one observer sees a vertical path, the other a nearly horizontal path.

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