PHY 150 Project One Supply Drop Plan Template_KAbbott
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Southern New Hampshire University *
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Subject
Industrial Engineering
Date
Apr 3, 2024
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docx
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A&L ENGINEERING
Supply Drop Plan
Keon Abbott
November 9
th
, 2023
PHY 150-J2326
3-2 Project One: Objects in Motion
Scenario
You have recently joined the team at A&L, an engineering firm with a broad portfolio. A&L has recently been hired to help plan a supply drop following a natural disaster. Due to conditions on the ground, the supply drop will be done from the air. Your supervisor has asked you to use that information, as well as your knowledge of kinematics, to create a supply drop plan detailing how
far the payload should be from the drop site when it is delivered. Additionally, due to adverse conditions in the area, your supervisor has asked you to prepare for two contingencies.
Directions
The plane carrying the supplies will be cruising at a constant velocity of 250 miles per hour relative to the ground and at a height of 2,650 meters above the target site. Using this information, create a supply drop plan
including all required information and calculations outlined below. As you are completing your supply drop plan, remember that correct SI units are a required component of your calculations and descriptions.
Horizontal and Vertical Motion
Diagram
*Can be seen on next page
Kinematics Calculations
*can be seen on next page
𝑦
=
𝑦
𝑜
+
𝑣
𝑜
𝑡
+
1
2
𝑔𝑡
2
0
=
−
2,650
+
4.9
𝑡
2
2,650
=
4.9
𝑡
2
2,650
4.9
=
𝑡
2
√
2,650
4.9
=
𝑡
𝑡
=
23.26
𝑠
Initial Velocity
Horizontal: 250 mi/h (or 402 km/h)
Vertical: 0
Final Velocity
Horizontal: 250 mi/h (or 402 km/h)
Vertical: 141.64 mi/h (or 227.95 km/h)
𝑣
𝑦
=
𝑣
𝑜
−
𝑔𝑡
𝑣
𝑦
=
0
−
9.8
⋅
0.11
𝑣
𝑦
=
−
288
𝑥
=
𝑣
𝑜
𝑡
𝑥
=
23.26
⋅
0.11
(km/s)
𝑥
=
2.6
Km
𝐷
=
𝑉𝑇
𝐷
=
0.11
(km/s) 23.26(s) 𝐷
=
¿
2.56 km or 2,560 meters
Equation Descriptions
Time of travel was calculated by using this equation: 𝑦
=
𝑦
𝑜
+
𝑣
𝑜
𝑡
+
1
2
𝑔𝑡
2
Finding the horizontal distance traveled was calculated by using this equation: 𝑥
=
𝑣
𝑜
𝑡
The velocity on landing was calculated by using this equation: 𝑣
𝑦
=
𝑣
𝑜
−
𝑔𝑡
The distance needed to travel to drop the load at the accurate spot was calculated by using this equation: 𝐷
=
𝑉𝑇
To truly understand this calculation, it is imperative to know that the horizontal motion will always stay consistent, because there is no acceleration. Vertical velocity is different in this scenario because gravity is responsible for the force on the object. There is no initial vertical velocity which results in the final velocity being negative. This is an
accurate answer because the object (cargo) is being dropped from the plane. This is when I
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made my calculations on when the load should be dropped. By doing so, this gave me the answer of the load should be dropped 2,560 meters (about the length of 30 city blocks) from the drop site in order to land accurately.
Adaptations for Scenario One
Diagram
*shown on next page
Adaptation Description
The current windspeed in Phoenix, Arizona sits at 3 miles per hour. 3 mph = 0.0013 km/s
To calculate the speed of the plane, we use this equation: v(plane) - v(windspeed)
0.11 km/s - 0.0013 km/s = 0.1087 km/s
To calculate when to drop the load to ensure an accurate arrival, we plug in the numbers:
0.1087 km/s * 23.26(s) = 2.52 km (or 2,520 meters). 2,520 meters (about the height of Mount St. Helens) is the length needed to drop the crate, and this is a difference of a mere 40 meters with the windspeed. 40 meters is as long as three school buses, so it is not much of a margin in
distance. However, if the plane was to fly in the middle of our monsoon season, it would be an
entirely different story. Adaptations for Scenario Two
Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for slinging supplies to affected areas, similar to the electromagnetic lift catapults used to launch planes from aircraft carriers. This catapult is located at a fixed point 400 meters away and 50 meters below the target site. The catapult is capable of launching the payload at 67 meters per second and an initial launch angle of 50 degrees. Using your knowledge
of kinematics equations, determine whether this would be sufficient to deliver the payload to the drop site.
Diagram
Adaptation Description
With this scenario, an entirely different approach needs to be used. This is essentially a little bit different in terms of having velocity from both the force of the catapult launching and the force of the wind/gravity that takes hold of the object being tossed in the sky. We will be using trigonometry to be able to solve whether the package will be launched high enough to reach the target. As you can see in the equation below:
cos50
=
𝑣
𝑥
67
67
(
cos50
)
=
𝑣
𝑥
𝑣
𝑥
=
43.07
m/s
sin 50
=
𝑣
𝑦
67
67
(
sin 50
)
=
𝑣
𝑦
𝑣
𝑦
=
51.33
400
43.07
=
9.29
seconds
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(
51.33
⋅
9.29
)
+
(
−
9.8
⋅
9.29
⋅
9.29
2
)
= 53.97 feet
With this answer, we know that with the calculations of 53.97 feet tells us that it will be able to launch high enough to reach its destination at the top. While working on this assignment, it was clear to me that the calculations depend on the variables. The results can instantly change if any of the variables change. There was a clear conclusion between where the crate would have landed if the variable in wind speed was different.
The scenario with the catapult was entirely different, quite the opposite I must say. It included a different set of equations to come to an answer. It also included a different method of acceleration, because it was getting launched and not dropped.
References
1.
Contributors to Lion King/Lion Guard Wikia. (n.d.). Pride rock
. Lion King/Lion Guard Wikia. https://lion-kinglion-guard.fandom.com/wiki/Pride_Rock
2.
Phoenix, AZ weather forecast and conditions - the weather channel
. The Weather Channel. (n.d.). https://weather.com/weather/today/l/Phoenix+AZ?
canonicalCityId=2e5ae66bd2b43b15b0bfc4f6a3628b6e13228e32d0a57fb643144293f78c
a339
3.
Urone, P. P., & Hinrichs, R. (2022). College physics
. OpenStax. 4.
Wikimedia Foundation. (2023, October 9). Grand Canyon National Park
. Wikipedia. https://en.wikipedia.org/wiki/Grand_Canyon_National_Park