## Statement

Quarantine was declared in a passenger cruise ship due to a contagious viral intoxication. To help, the Red Cross sent a helicopter with a box full of drugs. Since the crew of the helicopter could not land on the ship, it is decided to drop the package on a mat on the ship deck.

Assuming that the cruise travels at 72 km/h and that the helicopter travels in the same direction at 108 km/h, at an altitude of 40 m, at what horizontal distance from the ship should the package be dropped? And at what distance if they are traveling toward each other?

## Solution

**Data**

- Ship velocity: 72 km/h = 20 m/s
- Helicopter velocity: 108 km/h = 30 m/s

**Preliminary considerations**

- The ship is moving with uniform rectilinear motion
- The package will experience horizontal launch motion
- We will place the origin of the system of reference at sea level, with the y-axis at the point at which the helicopter will drop the package
- At that time, the ship is at a distance
*x*which is exactly what we are being asked to calculate_{0s }

Starting from these considerations above, the data of the motion are:

- Initial velocity of the package in the
*x-axis: v*= 30 m/s_{0xp} - Initial velocity of the package in the
*y-axis: v*= 0 m/s_{0yp} - Ship velocity in the
*x-axis: v*= 20 m/s_{0xs} - Initial height of the package:
*y*=_{0p}*H*= 40 m - Final height of the package:
*y*= 0 m_{fp} - Value of the acceleration of gravity
*g*= 9.8 m/s^{2}

**Resolution**

The equations for each motion are shown in the following table. Notice we use the subscripts *p* and *s* to refer to the package and ship variables respectively.

Motion of the package | Motion of the ship | |
---|---|---|

x-axis |
$${x}_{{f}_{p}}=\overline{){x}_{{0}_{p}}}+{v}_{0{x}_{p}}\xb7t$$ | $${x}_{{f}_{s}}={x}_{{0}_{s}}+{v}_{0{x}_{s}}\xb7t$$ |

y-axis |
$${y}_{{f}_{p}}={y}_{{o}_{p}}+\overline{){v}_{0{y}_{p}}\xb7t}-\frac{1}{2}\xb7g\xb7{t}^{2}$$ $${v}_{f{y}_{p}}={v}_{0{y}_{p}}-g\xb7t$$ | - |

Notice that we have followed the usual sign convention for linear motion.

The condition for the package to fall on the ship is matching the coordinates *x* and *y* from the two motions at time *t*. That is:

$${x}_{{f}_{p}}={x}_{{f}_{s}}\phantom{\rule{0ex}{0ex}}{y}_{{f}_{p}}={y}_{{f}_{s}}=0$$

Therefore, let us first look at the *y* coordinate, it have to be 0 since we are considering the landing point to be at sea level. That way we can determine the time that the package is in the air.

$${y}_{{f}_{p}}={y}_{{o}_{p}}-\frac{1}{2}\xb7g\xb7{t}^{2}\Rightarrow 0=40-\frac{1}{2}\xb79.8\xb7{t}^{2}\Rightarrow t=2.85m/s$$

On the other hand, knowing that the *x* coordinates must also match, we determine the horizontal distance at which the package must be released:

$${x}_{{f}_{p}}={x}_{{f}_{s}}\Rightarrow {v}_{0{x}_{p}}\xb7t={x}_{{0}_{s}}+{v}_{0{x}_{s}}\xb7t\Rightarrow {x}_{{0}_{s}}=t\xb7({v}_{0{x}_{s}}-{v}_{0{x}_{s}})=2.85\xb7(30-20))=\overline{)28.5m}$$

As for the second question, what would happen if the ship and helicopter were traveling toward each other (they go in opposite directions), simply we change the sign of *v _{0xs}*, to

*v*= -20 m/s. That way we satisfy the sign convention mentioned earlier.

_{0xs}$${x}_{{f}_{p}}={x}_{{f}_{s}}\Rightarrow {v}_{0{x}_{p}}\xb7t={x}_{{0}_{s}}+{v}_{0{x}_{s}}\xb7t\Rightarrow {x}_{{0}_{s}}=t\xb7({v}_{0{x}_{p}}-{v}_{0{x}_{s}})=2.85\xb7(30-(-20\left)\right)=\overline{)142.5m}$$

Indeed, since the ship and the helicopter go toward each other, it is necessary that the package be released at a greater distance.

One last note. Remember that what we have calculated is the *horizontal* distance between the ship and the helicopter at the time that the package is released, that is, the distance between the ship and the vertical line in which the helicopter is at the time that the package is released. If we were only asked for the distance, we must consider the *y* coordinate as well. That is:

$$d=\sqrt{{x}^{2}+{y}^{2}}$$

Therefore, in the first case:

$$d=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{28.{5}^{2}+{40}^{2}}=49.11m$$

In the second case:

$$d=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{142.{5}^{2}+{40}^{2}}=148.08m$$