Firing cannons projectile motion

In summary, we are given a cannonball that is fired with an initial velocity of v0 at an angle of 30 degrees above the horizontal from a height of 38.0 m above the ground. The cannonball strikes the ground with a speed of 1.3v0. Using the kinematic equations, we can find the maximum altitude reached by the cannonball and use the conservation of energy to determine the magnitude of the initial velocity. From there, we can use the components of velocity to find the magnitude of the initial velocity.
  • #1
Psychros
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Homework Statement


A cannonball is fired with initial speed v0 at an angle 30° above the horizontal from a height of 38.0 m above the ground. The projectile strikes the ground with a speed of 1.3v0. Find v0. (Ignore any effects due to air resistance.)
ScreenShot2014-04-26at71948PM.png


Homework Equations


Time= 0=v0T-½gT2 T<0
t= (vOy+√(vOy2-2gy))/g
distance/range x=x0+V0xt
x/y velocity- v0x=Vcos(θ); v0y=vsin(θ)


The Attempt at a Solution



This is a practice problem (with different values than original), and the provided answer is 32.9m/s

i set the origin at the cannon
Since the velocity magnitude = √(Vcos(θ)2+Vsin(θ)2)

I tried variations along the line of V0cos(30)2+V0sin(30)2= 1.3 V02
in order to split the velocity into xy components.

Conceptually i understand it takes him -38m to reach 1.3 times his original velocity, but in trying to plug the provided data into the equations i end up with too many variables.

Any and all help very appreciated, thank you.
 
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  • #2
Psychros said:

The Attempt at a Solution



This is a practice problem (with different values than original), and the provided answer is 32.9m/s

i set the origin at the cannon
Since the velocity magnitude = √(Vcos(θ)2+Vsin(θ)2)

I tried variations along the line of V0cos(30)2+V0sin(30)2= 1.3 V02
in order to split the velocity into xy components.

Conceptually i understand it takes him -38m to reach 1.3 times his original velocity, but in trying to plug the provided data into the equations i end up with too many variables.

Any and all help very appreciated, thank you.

I'm not sure what you are trying to do here. The angle at which the projectile strikes the ground may not be equal to the initial firing angle, i.e. 30 degrees.

The best way to approach this problem is to find the maximum altitude which the projectile reaches after it is fired. At this point, the vertical component of the velocity is zero. After reaching the maximum altitude, the projectile falls under the force of gravity. Knowing the magnitude of the final velocity, you should be able to determine the duration of the fall and the resulting vertical velocity component. Remember, since there is no air resistance, the horizontal component of the velocity remains unchanged during the entire flight of the projectile.
 
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  • #3
SteamKing said:
I'm not sure what you are trying to do here. The angle at which the projectile strikes the ground may not be equal to the initial firing angle, i.e. 30 degrees.

The best way to approach this problem is to find the maximum altitude which the projectile reaches after it is fired. At this point, the vertical component of the velocity is zero. After reaching the maximum altitude, the projectile falls under the force of gravity. Knowing the magnitude of the final velocity, you should be able to determine the duration of the fall and the resulting vertical velocity component. Remember, since there is no air resistance, the horizontal component of the velocity remains unchanged during the entire flight of the projectile.

That is brilliant, thank you. However, Without knowing distance (x), time, or hard value for velocity (since it's expressed as a relation to the initial velocity) I'm not sure how to go about this.
 
  • #4
Psychros said:

Homework Statement


A cannonball is fired with initial speed v0 at an angle 30° above the horizontal from a height of 38.0 m above the ground. The projectile strikes the ground with a speed of 1.3v0. Find v0. (Ignore any effects due to air resistance.)
ScreenShot2014-04-26at71948PM.png


Homework Equations


Time= 0=v0T-½gT2 T<0
t= (vOy+√(vOy2-2gy))/g
distance/range x=x0+V0xt
x/y velocity- v0x=Vcos(θ); v0y=vsin(θ)
...
Try using conservation of energy.
 
  • #5
SammyS said:
Try using conservation of energy.

Hmmm I don't think we've learned that yet..
 
  • #6
Psychros said:
Hmmm I don't think we've learned that yet..
In that case,

What can you conclude about the x component of the velocity?


What kinematic equations do you know regarding the vertical component of the velocity ?
 
  • #7
X component will be the same through it's entirety.

y=y0+v0y(t)+½at2
or
Vy=V0y+gt
Vy=1.3V0y-9.81t
But Theoretically if we were starting at the maximum, V0y would = 0,

Vy/-9.81=t?

I'm getting stuck down either avenue, I do think I'm missing something intuitive here..
 
  • #8
Psychros said:
X component will be the same through it's entirety.

y=y0+v0y(t)+½at2
or
Vy=V0y+gt
1.3Vy=V0y-9.81t
1.3Vy=0-9.8t
Vy=(-9.8t)/1.3

I'm getting stuck down either avenue, I do think I'm missing something intuitive here..
vy ≠ 1.3 v0y . I'm assuming that you mean vy to be the y component of the final velocity.

How is speed obtained from components of velocity ?


Also, there's another kinematic equation. One that doesn't involve time, t.

Edited:
[STRIKE](vy)2 = (v0y)2 - g(Δy)[/STRIKE]

(vy)2 = (v0y)2 - 2g(Δy)
 
Last edited:
  • #9
Oh yes, i made a false assumption about the y component.
Do you mean speed as in magnitude of velocity?

Also, should (vy)2 = (v0y)2 - g(Δy)
have a -2g(Δy)?

I apologize for being slow, but I'm not sure how to use this either since I don't know the change in height from the maximum
 
  • #10
Psychros said:
Oh yes, i made a false assumption about the y component.
Do you mean speed as in magnitude of velocity?

Also, should (vy)2 = (v0y)2 - g(Δy)
have a -2g(Δy)?

I apologize for being slow, but I'm not sure how to use this either since I don't know the change in height from the maximum
Right. There should be a 2 in there.


Then, ...

I'll ask again,
How is speed obtained from components of velocity ?​
 
  • #11
Since speed is the magnitude of velocity,
by components of velocity I imagine you mean Vx and Vy,
or √(Vcos(θ)2+Vsin(θ)2)?
 
  • #12
Psychros said:
Since speed is the magnitude of velocity,
by components of velocity I imagine you mean Vx and Vy,
or √(Vcos(θ)2+Vsin(θ)2)?
Yes. Of course, sin2(θ) + cos2(θ) = 1

Anyway, ##\ v_0=\sqrt{(v_{0x})^2+(v_{0y})^2}\ ,\ ## so that ##\ (v_0)^2=(v_{0x})^2+(v_{0y})^2\ .\ ##

And in general ##\ v^2=(v_{x})^2+(v_{y})^2\ .\ ## Right?


Now, what do you know regarding the x component of velocity for this projectile?
 
  • #13
sorry, that's what i meant when i put √(V0cos(θ)2+V0sin(θ)2), that it's = V0.

I know that it doesn't change, but I'm not sure how to find its' magnitude in this situation.
 
  • #14
Psychros said:
sorry, that's what i meant when i put √(V0cos(θ)2+V0sin(θ)2), that it's = V0.

I know that it doesn't change, but I'm not sure how to find its' magnitude in this situation.

So then ##v_{0x}=v_x\ ## .

Regarding your kinematic equation with the ##(v_{y})^2\, ##, what is it missing for it to have v2 and v02 in it ?
 

FAQ: Firing cannons projectile motion

What is the relationship between the angle of a cannon and the distance traveled by the projectile?

The angle of a cannon affects the distance traveled by a projectile due to the laws of projectile motion. Generally, the greater the angle of the cannon, the greater the distance the projectile will travel. However, there is an optimal angle that will result in the maximum distance traveled, known as the angle of projection.

How does the weight of a cannonball affect its trajectory when fired from a cannon?

The weight of a cannonball does not significantly affect its trajectory when fired from a cannon. This is because in projectile motion, the vertical and horizontal components of motion are independent of each other. The weight of the cannonball only affects the vertical motion, while the horizontal motion is determined by the initial velocity and angle of projection.

What is the role of air resistance in the trajectory of a projectile fired from a cannon?

Air resistance does play a role in the trajectory of a projectile, but it is typically very small compared to other factors such as gravity and initial velocity. The effect of air resistance can be reduced by using a streamlined projectile or firing at higher velocities.

How does the height of a cannon affect the range of the projectile it fires?

The height of a cannon does not have a significant effect on the range of the projectile it fires. As long as the cannon is positioned at ground level, the height does not change the initial velocity or angle of projection, which are the main factors that determine the range of the projectile.

What are some real-world applications of understanding projectile motion in firing cannons?

Understanding projectile motion in firing cannons has many practical applications, such as in military operations, sporting events, and even space exploration. It can also be used in ballistics analysis and the design of trajectory-based weapons. Additionally, understanding projectile motion can help engineers design better and more efficient cannons and other ballistic devices.

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