In this article (and video above), we dive deep into the fascinating world of projectile motion and explore how to determine the exact position of a projectile at its maximum height. This question forms part of the Dynamics section in the FE exam. Join us as we break down the principles of kinematics and the fundamental concepts of projectile motion, including the independence of horizontal and vertical components when air resistance is negligible
Question:
A projectile is shot from ground level with an initial velocity of 22 m/s at an angle of 40°. Determine its (x,y) coordinate when it reaches its peak altitude. Neglect air resistance.
Problem Context:
Projectile motion
The motion of an object thrown or projected into the air, subject to only the acceleration of gravity. (OpenStax, Rice University)
Kinematics of particles
The study of the movement of particles, without considering the forces that cause this movement. (Collins Dictionary)
The equations governing projectile motion are derived from the well-established constant acceleration equations.
- No air resistance: horizontal and vertical motions independent
- Horizontal motion of a projectile is constant (ax=0)
- Vertical motion of a projectile is influenced by gravity (ay=-g)
These components are the key to deriving the projectile motion equations specific to each dimension. As mentioned before, these equations account for the constant velocity in the horizontal direction and the gravitational acceleration in the vertical dimension. These are the equations we will be applying to analyze our problem. So, without further ado, let’s jump right into the solution!
Step 1 – Time at maximum height
Calculate the time to reach maximum height:
At the maximum height, the vertical velocity v_y=0 m/s
Step 2 – Displacement
1. Calculate the maximum height:
Knowing the time it takes for the projectile to reach its highest point, we can now determine its vertical position, or y-coordinate, at this moment. When inspecting the formulas we have available, we see that we have all the necessary information to find this value using equation 4
We can directly substitute our known time, initial velocity and launch angle to find that its vertical position at this point is equal to 10.19 metres.
2. Calculate the x-coordinate at maximum height:
To complete our analysis, we’ll determine the horizontal position, or x-coordinate, using equation 2.
As before, we substitute our known values and find that the x-coordinate is equal to 24.29 meters.
This concludes our analysis, yielding the final (x, y) coordinates of the projectile at its maximum height.
Answer:
The correct answer is D.
Final Remarks:
1.Horizontal and vertical motion is only independent when air resistance is negligible.
2.A projectile only experiences acceleration in the vertical direction (ay=-g).
3.A projectile’s vertical velocity is zero at its highest point.
4.Understanding = easy application in any scenario.
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I hope you found this article helpful. In upcoming articles, I will solve some more PE exam practice problems and answer other questions from our subscribers. Pass the PE Exam videos will publish weekly, so be sure to click the subscribe button so you don’t miss something that could make a substantial difference in your exam result.
Lastly, I encourage you to ask questions in the comments of this video, or on this page and I’ll read and respond to them in future videos. So, if there’s a specific topic you want me to cover or answer, we have you covered.
I’ll see you next week… on Pass the FE Exam
Anthony Fasano, P.E.
Engineering Management Institute
Author of Engineer Your Own Success
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