# Projectile Trajectory

1. Aug 30, 2015

### Beanie

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

d=Vit+1/2at^2

3. 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.

2. Aug 30, 2015

### Bystander

3. Aug 30, 2015

### Beanie

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

4. Aug 30, 2015

### Bystander

No idea how to find components of a vector?

5. Aug 30, 2015

### SteamKing

Staff Emeritus
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.

6. Aug 31, 2015

### Beanie

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?

7. Aug 31, 2015

### Beanie

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?

8. Aug 31, 2015

### SteamKing

Staff Emeritus
Did you remember what I told you about gravity? What goes up must come back down?

9. Aug 31, 2015

### Beanie

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. Aug 31, 2015

### SteamKing

Staff Emeritus
Yes.

11. Aug 31, 2015

Thank you!