Calculating Bullet Trajectory Distance at 30° Angle

[Robert Alan]

Homework Statement

A bullet is shot at initial velocity of 600m/s at an angle of 30 degrees. How far will it travel before landing? (Assume no air resistance)
v=600m/s
a=9.8m/s^2
t=?
distance=?

Homework Equations

x=x0+v0t+1/2at^2

The Attempt at a Solution

I'm stuck on this one. I can calculate its distance shot straight up, or from a level surface and I can calculate its time shot straight up or from a level surface. The angle is throwing me off. Does this have something to do with a cosine of 30 degrees?

 

[diazona]

Yep To start, split the motion into horizontal and vertical components.

 

[Robert Alan]

Okay. The horizontal component at 30 degrees is = 600cos30? And the vertical component is 600sin30? What now?

 

[rock.freak667]

Okay. The horizontal component at 30 degrees is = 600cos30? And the vertical component is 600sin30? What now?

Use the vertical components to find the total time taken for the motion


$$x=x_0 + v_0t - \frac{1}{2}gt^2$$

 

[Robert Alan]

Okay. So I get x=300t-0.5t^2 (For this problem we assume g=10m/s^2). At 30 seconds it will attain a height of 4500. I got this by putting y=300t-5t^2 in my calculator and finding the maximum. Now I know how high it goes and how long it is in the air. Now what do I do to find the range of it?

 

[diazona]

Use the horizontal velocity to find out how far the bullet travels during the time it is in the air.

 

[rock.freak667]

Okay. So I get x=300t-0.5t^2 (For this problem we assume g=10m/s^2). At 30 seconds it will attain a height of 4500. I got this by putting y=300t-5t^2 in my calculator and finding the maximum. Now I know how high it goes and how long it is in the air. Now what do I do to find the range of it?

When the bullet hits the ground, the displacement is zero, so y=0 and find t.

 

[Robert Alan]

Oh snap! So I got 30sec at the max time, double it because the rise and fall are symmetrical. That gives 60sec. Then using x=vt or x=600cos30(60) I got 31176.91 meters! Is that right?

 

[rock.freak667]

that should be correct.

 

[Robert Alan]

Thanks so much for the help! I'm starting to see things now in terms of vectors. Very interesting! :)

 

[MLeszega]

Okay. So I get x=300t-0.5t^2 (For this problem we assume g=10m/s^2). At 30 seconds it will attain a height of 4500. I got this by putting y=300t-5t^2 in my calculator and finding the maximum. Now I know how high it goes and how long it is in the air. Now what do I do to find the range of it?

I am pretty bad at physics so I am trying to follow the problem but I get stuck here. I thought that v₀=600m/s. How come you wrote 300? Is it because v₀ is m/s while gravity is m/s²?

 

[rock.freak667]

I am pretty bad at physics so I am trying to follow the problem but I get stuck here. I thought that v₀=600m/s. How come you wrote 300? Is it because v₀ is m/s while gravity is m/s²?

Vertically for x=x₀ +v₀t-(1/2)gt²

v₀ is the vertical component of the initial velocity.

 

[MLeszega]

Vertically for x=x₀ +v₀t-(1/2)gt²

v₀ is the vertical component of the initial velocity.

So v₀ = 600sin(30) for the vertical component and 600cos(30) for the horizontal component?

 

[rock.freak667]

So v₀ = 600sin(30) for the vertical component and 600cos(30) for the horizontal component?

yes.

 

[MLeszega]

yes.

Ahh, ty very much!