Artillery shell time of flight at 23.9° and 1530 m/s

[Beanie]

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.

 

[Bystander]

I don't know final velocity

23.9◦ above the horizontal ground with an initial speed of 1530 m/s.

neglecting air resistance.

 

[Beanie]

I understand this, however how can you calculate final velocity with just the angle and initial velocity?

 

[Bystander]

No idea how to find components of a vector?

 

[SteamKing]

I understand this, however how can you calculate final velocity with just the angle and initial velocity?

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.

 

[Beanie]

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.

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?

 

[Beanie]

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?

 

[SteamKing]

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?

 

[Beanie]

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?

 

[SteamKing]

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.

 

[Beanie]

Yes.

Thank you!