Projectile Trajectory Homework: Finding Time of Flight Without Air Resistance

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Homework Help Overview

The discussion revolves around calculating the total time of flight for a projectile, specifically an artillery shell fired at an angle of 23.9° with an initial speed of 1530 m/s, while neglecting air resistance. The problem is situated within the context of projectile motion and kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the challenge of determining the final velocity with only the initial velocity and angle provided. There are attempts to break down the velocity into components and to apply kinematic equations, but confusion arises regarding the correct application of these concepts.

Discussion Status

Some participants have provided guidance on visualizing the projectile's trajectory and have suggested drawing diagrams to aid understanding. There is an ongoing exploration of the relationship between the vertical and horizontal components of motion, and some participants are questioning their calculations and assumptions regarding time and velocity.

Contextual Notes

Participants express uncertainty about the correct use of kinematic equations and the implications of gravity on the projectile's motion. There is a recognition that the time calculated may only represent half of the total flight time, prompting further discussion on how to adjust their calculations accordingly.

Beanie
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Homework Statement


An artillery shell is fired at an angle of 23.9◦ above the horizontal ground with an initial speed of 1530 m/s.

The acceleration of gravity is 9.8 m/s2 .

Find the total time of flight of the shell, neglecting air resistance.

Answer in units of min.

Homework Equations



Vf^2=Vi^2+2ad
d=Vit+1/2at^2

The Attempt at a Solution


I tried using the first equation above with the givens to find distance, however, I don't know final velocity. Once finding the distance I would have plugged it into the second equation above to find the time. I would have then converted the time in seconds to find the time in minutes.

However, none of this was possible because I didn't know the final velocity.
 
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Beanie said:
I don't know final velocity

Beanie said:
23.9◦ above the horizontal ground with an initial speed of 1530 m/s.
Beanie said:
neglecting air resistance.
 
I understand this, however how can you calculate final velocity with just the angle and initial velocity?
 
No idea how to find components of a vector?
 
Beanie said:
I understand this, however how can you calculate final velocity with just the angle and initial velocity?

Beanie said:
Find the total time of flight of the shell, neglecting air resistance.

Since you are firing the projectile at an elevated angle above the horizontal, what happens to the projectile once it's fired? You can draw a picture of the trajectory if that will help you to visualize what's going on.

Remember, since you are dealing with gravity, what goes up must come back down. :wink:
 
SteamKing said:
Since you are firing the projectile at an elevated angle above the horizontal, what happens to the projectile once it's fired? You can draw a picture of the trajectory if that will help you to visualize what's going on.

Remember, since you are dealing with gravity, what goes up must come back down. :wink:
Thank you! So I understand what the projectile looks like. I have drawn a picture and tried a different attempt to what I was doing before. I found the x and y components of the angle. I then used the y component of this answer as the initial vertical velocity, used 0 as the final vertical velocity, and used-9.8 as the vertical acceleration. I then plugged these values into the Vy=Viy+at to find the time in seconds. I got 63.25s. I then converted this to 1.054 minutes. However, this answer was wrong.

I still don't understand what I am doing wrong. Any ideas?
 
Bystander said:
No idea how to find components of a vector?

Yes, I used the y component of the vector as explained in my reply to "SteamKing" However, I still got the answer wrong.

Any ideas on where I went wrong?
 
Beanie said:
Thank you! So I understand what the projectile looks like. I have drawn a picture and tried a different attempt to what I was doing before. I found the x and y components of the angle. I then used the y component of this answer as the initial vertical velocity, used 0 as the final vertical velocity, and used-9.8 as the vertical acceleration. I then plugged these values into the Vy=Viy+at to find the time in seconds. I got 63.25s. I then converted this to 1.054 minutes. However, this answer was wrong.

I still don't understand what I am doing wrong. Any ideas?
Did you remember what I told you about gravity? What goes up must come back down?
 
SteamKing said:
Did you remember what I told you about gravity? What goes up must come back down?
Oh right, okay so my time is half of the time that it takes the shell to fly. So I must double my time?
 
  • #10
Beanie said:
Oh right, okay so my time is half of the time that it takes the shell to fly. So I must double my time?
Yes.
 
  • #11
SteamKing said:
Yes.

Thank you!
 

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