**Ballistic trajectory or the Black Lagoon Helicopter stunt**

Welcome class and please get out your physics books as this week’s lesson is on trajectory, specifically ballistic trajectory and Black Lagoon. No we will not be examining bullet trajectories from the numerous shootouts found in the show, but the infamous PT boat vs Helicopter scene. For those of you who are not familiar with it, the clip can be found below.

**Considerations**

Now just at first glance, there are several factors that need to be examined, to determine if this stunt is possible in real life. The speed of the boat, angle of the ramp, height of the helicopter, not to mention the strength of the ramp, and the strength of the boat’s hull. For the purposes of this examination I will not be including the strength of the ramp or the strength boat’s hull. The reason for this exclusion is because a PT boats hull is made out of wood, specifically mahogany, which while strong, does have its limits. This is also the reason for excluding the strength of the sunken boat that is used as a ramp. From the few shots that are shown of the boat it sports a fair amount of rust, and has been sitting there for an undetermined length of time. Given the amount of force needed to launch a boat into the air, it is entirely possible that the PT boats hull or the ramp could give out as soon as they collide ending the myth right there. Now given that the ramp is successfully used in the clip, we will assume that both the hull of the boat and the ramp are strong enough.

**Terminology**

In order to determine if the Black Lagoon can attack the helicopter with torpedoes, the trajectory of the Black Lagoon must be determined. Trajectory is the path a projectile takes through space. While the Black Lagoon is a PT and not a true projectile, once it leaves the water and can no longer propel itself, it can be considered one. The type of trajectory that we are interested in is a ballistic trajectory. In this type of motion the object moves both vertically and horizontally near the earth’s surface. The force of gravity is the main force acting on the projectile so other forces like air resistance are excluded as their influence in this case would be negligible at best.

**Basic Facts of the scene**

Before we begin diving into the math let’s consider the facts that will be used to consider if this particular scene is in fact possible. In the picture below we can see that at the peak of its ballistic trajectory the Black Lagoon reaches a height greater than that of the helicopter.

We can use a second picture to estimate the approximate altitude of the helicopter.

In this picture we are concerned about the trees shown in the background as a way to determine the approximate altitude. We can assume at this point in the show that the Black Lagoon is in an estuary close to Ronanapur, a fictional city in Thailand. Mangrove trees are a common tree found in estuaries in Thailand. There are several different species mangroves with a range of heights from 6 meters on the shorter side to species that can grow up to 35 meters on the larger side. Now let’s assume that trees beneath the helicopter are of the shorter variety at 6 meters. An examination of the image shows that the helicopter is two tree lengths from the water at a height of 12 meters. Given the first image the Black Lagoon must reach a height of at least 12 meters for the scene to be possible.

**The Math**

Projectile motion also known as ballistic trajectory is motion in two directions and the equations for each are shown below.

**Glossary**

**x-** Horizontal position

**y-** Vertical position

**v-** velocity- speed

**t-** Time

**a-** acceleration- change in speed= to the force of gravity at 9.8m/s^{2}

**f-** Final

**i-** Initial

Determining horizontal position

** X _{f} = vt + X_{i}**

Determining vertical position

** Y _{f} = Y_{i} +v_{i}t_{i} +at_{f}^{2}**

**V _{f}^{2} = V_{i}^{2} +2a(Y_{f} – Y_{i})**

The above equations work in situations where the object is not projected at an angle like a plate sliding off the edge of a table. If the projectile is launched at an angle another wrinkle needs to be added in order to complete the calculations for a ballistic or parabolic trajectory shown below.

As shown in the diagram the angle of the launch determines the height and distance the projectile travels assuming each one is launched with the same initial speed. For the purposes of this evaluation only the vertical component of the Black Lagoons ballistic trajectory will be considered as it does fly past the helicopter in the scene, and they can be considered close enough for the torpedoes to hit the target. The maximum height will be determined for each of the three angles shown above

Calculating the vertical component of the initial velocity

**V _{yi} = V_{xi}(sin angle)**

The maximum speed of a PT boat is 76kmph and it can be assumed that Dutch keeps the Black Lagoon in top form. To make the math easier 76kmph converts to 21.1 meters per second.

Vertical speed at 30 degrees

** V _{yi}** = 21.1m/s x sin30

** V _{yi} **= 21.1 x 0.5

Initial vertical velocity at 30 degrees = 10.6 m/s

Vertical speed at 45 degrees

** V _{yi} **= 21.1m/s x sin45

** V _{yi}** = 21.1m/s x 0.71

Initial vertical velocity at 45 degrees = 14.9 m/s

Vertical speed at 60 degrees

** V _{yi} **= 21.1m/s x sin60

**V _{yi} **= 21.1m/s x 0.87

Initial vertical velocity at 60 degrees = 18.4 m/s

Calculating the maximum height of the Black Lagoon

**Maximum height = (V _{yi}^{2}) / (2a)**

At 30 degrees

X= (10.6)^{2} / 2 x 9.8

X = 112.4 / 19.6

Max height = 5.7 meters

At 45 degrees

X= (14.9)^{2} / 2 x 9.8

X = 222 / 19.6

**Max height = 11.32 meters**

At 60 degrees

X= (18.4)^{2} / 2 x 9.8

X = 338.6 / 19.6

** Max height = 17.3 meters**

**Conclusion**

Well the numbers don’t lie guys, and to be honest I was surprised by this one. If the angle of the sunken boat the Black Lagoon uses as a ramp is more than 45 degrees then the ballistic trajectory of the boat will reach the 12 meter mark and the stunt works. I did not expect to be able to call this one confirmed but here it is, PT boat vs helicopter is

The infamous PT boat vs helicopter scene can actually work if we assume the hull of the boat and the ramp can survive the stress. Plus Dutch has to be able to aim the torpedoes to hit the helicopter. If you have any comments or suggestions, please leave them in the comments below.

**Sources**

http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

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