My professor's crazy cannon question HELP please

  • Thread starter Cassomophone
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In summary: Vx)i Δt=(Vy)iΔt-gΔt so...Vx and Vy are related by Vx=(Vy)i-gΔt. So you could try substituting Vx into eqn. 1) and see if that gives the correct answer. or you could try substituting Vy into eqn. 3) and see if that gives the correct answer.
  • #1

Homework Statement



So here is the question my professor made up. I asked if the time was missing and once he thought about it, he said we should be able to figure it out. I need help.

A cannon is mounted on a railroad track on a cliff above the ocean. 300 m below, and standing off a 15 km, is a war ship.The cannon's elevation is fixed and designed to provide maximum range (which I know to be 45°).

a) What is the required muzzle velocity of the cannon?

b) The ship moves with it's range, (It moves to 5km from cliff), so the cannon is rolled back on the track and the projectile charge changed, doubling the cannon's muzzle velocity. How far away does the cannon need to be rolled back to target the ship?

Homework Equations



Xf, Yf =X,Y final
Xi, Yi =X,Y initial
(Vx)i, (Vy)i= initial velocity of X, Y
(Vx)f, (Vy)f =final velosity of X,Y
g=gravity of 9.8 m/s^2
t=time
Vm=Muzzle Velocity

Equations (Eqn)
1) Xf=Xi+(Vx)i Δt
2) (Vx)f=(Vx)i= constant
3) Yf=Yi+(Vy)iΔt-(1/2)(g)(Δt)^2
4) (Vy)f=(Vy)i-gΔt
5) Vx+Vy=Vm

The Attempt at a Solution



So first I drew out the problem, I have the cannon on top of the 300m (which equals 3km) cliff and the ship is 15 km out. Then at the muzzle of the cannon I put an angle of 45°. There will be an arch for the cannonball to hit the ship.

Part A
θ=45°
g=9.8 m/s^2
Xf=15km
Xi=0
Yf=0
Yi=3km
(Vx)f and (Vy)f=0
Δt=? (Initial time is 0)
(Vx)i or (Vf)i=?

So I need either the initial velocity of X or Y to solve for Δt, or I really need a Δt. I have no idea how for find the Δt with what I know. Since the arch of the cannonball is not a perfect triangle can I use Ax=Acosθ and Ay=Asinθ? If I can get help on the time then I know I can probably solve A.
Once I have the time I would:
-solve for (Vx)i using eqn. 1.
-Solve for (Vy)i using eqn. 4.
-Then I would add them both to get the muzzle velocity of the cannon

For B
So the muzzle velocity from part A would be doubled
The ship would be 5km from the cliff at this point.
So now I need to find the Xi, so how should I do that exactly?


Thank you so much, I tried to be as detailed as possible and that is all the information that was given for the question.
 
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  • #2
Without actually doing anything, 300 m is not 3 km. 3000 m = 3 km; so that would be an error right off the bat. Make sure you've converted all units appropriately (for example - I would make everything in m instead of km since acceleration due to gravity is often given in m/s etc.). I'll take a look at this in more detail and get back to you with more feedback.
 
  • #3
Ah true, I didn't mean 3 km, I meant to type .3km. But you are right I should convert the km into meters so, 15km=15,000m and 5km is 5,000 m.

Thank you so much for looking at this.
 
  • #4
Moreover, the muzzle velocity is a vector Vxi+Vyj. You need the muzzle speed, the magnitude of the velocity, which is Vm=√(Vx2+Vy2). Knowing the angle of the muzzle velocity, you can express Vx and Vy with Vm.

Eliminate the time: You can isolate Δt from eq 1) and then substitute into equation 3).

ehild
 
  • #5
ehild said:
Moreover, the muzzle velocity is a vector Vxi+Vyj. You need the muzzle speed, the magnitude of the velocity, which is Vm=√(Vx2+Vy2). Knowing the angle of the muzzle velocity, you can express Vx and Vy with Vm.

Eliminate the time: You can isolate Δt from eq 1) and then substitute into equation 3).

ehild

So I isolate the time from eq 1, but I still don't know the initial velocity of x, so how do I do that?
 
  • #6
What are the components of the initial velocity (the nuzzle velocity) related to the magnitude and angle? You know the angle the ball is shoot. Unknown is the magnitude, but you know the initial and final positions, so you can determine it. ehild
 
  • #7
Ok... I get that I can, but how do I do that then?
 
  • #8
Tsunoyukami said:
Without actually doing anything, 300 m is not 3 km. 3000 m = 3 km; so that would be an error right off the bat. Make sure you've converted all units appropriately (for example - I would make everything in m instead of km since acceleration due to gravity is often given in m/s etc.). I'll take a look at this in more detail and get back to you with more feedback.

Ok so look at this real quick.
Since the equation is at 45 ° doesn't (Vx)i = (Vy)i? So if that is true...
then equation 1 gives:
15000m=(Vx)i Δt
equation 3 gives:
0=300+(Vy)iΔt-(1/2)(g)(Δt)^2
but since (Vx)i = (Vy)i then (Vy)iΔt = (Vx)i Δt = 15 so:
0=300+15000-(1/2)(g)(Δt)^2
so
15300=(1/2)(g)(Δt)^2
so
Δt=sqrt(30600/g)
giving:
(Vx)i =15000/sqrt(30600/g) = about 268m/sec
So the muzzle velocity would be 536 m/s right?

Now I am not sure where to go on part B. All i know is the muzzle velocity is doubled so that should be 2(536)= 1072 m/s
Now θ=45°
g=9.8 m/s^2
Xf=x+5
Xi=0
Yf=0
Yi=3km or is this 0 now since it's further back from the cliff?
(Vx)f and (Vy)f=0
Δt=? (Initial time is 0)
(Vx)i and (Vf)i=536 m/s
Vm= 1072 m/s

So using
Xf=X_o + Vx_o (Change of T)
0=X+5 +536m/s (Change of T)

And now I just confused myself, am I on the right track?
 
  • #9
ehild said:
What are the components of the initial velocity (the nuzzle velocity) related to the magnitude and angle? You know the angle the ball is shoot. Unknown is the magnitude, but you know the initial and final positions, so you can determine it.


ehild

Could I do this instead?


Since the equation is at 45 ° doesn't (Vx)i = (Vy)i? So if that is true...
then equation 1 gives:
15000m=(Vx)i Δt
equation 3 gives:
0=300+(Vy)iΔt-(1/2)(g)(Δt)^2
but since (Vx)i = (Vy)i then (Vy)iΔt = (Vx)i Δt = 15 so:
0=300+15000-(1/2)(g)(Δt)^2
so
15300=(1/2)(g)(Δt)^2
so
Δt=sqrt(30600/g)
giving:
(Vx)i =15000/sqrt(30600/g) = about 268m/sec
So the muzzle velocity would be 536 m/s right?

Now I am not sure where to go on part B. All i know is the muzzle velocity is doubled so that should be 2(536)= 1072 m/s
Now θ=45°
g=9.8 m/s^2
Xf=x+5
Xi=0
Yf=0
Yi=3km or is this 0 now since it's further back from the cliff?
(Vx)f and (Vy)f=0
Δt=? (Initial time is 0)
(Vx)i and (Vf)i=536 m/s
Vm= 1072 m/s

So using
Xf=X_o + Vx_o (Change of T)
0=X+5 +536m/s (Change of T)

And now I just confused myself, am I on the right track?
 
  • #10
Cassomophone said:
Ok so look at this real quick.
Since the equation is at 45 ° doesn't (Vx)i = (Vy)i? So if that is true...
then equation 1 gives:
15000m=(Vx)i Δt
equation 3 gives:
0=300+(Vy)iΔt-(1/2)(g)(Δt)^2
but since (Vx)i = (Vy)i then (Vy)iΔt = (Vx)i Δt = 15 so:
0=300+15000-(1/2)(g)(Δt)^2
so
15300=(1/2)(g)(Δt)^2
so
Δt=sqrt(30600/g)
giving:
(Vx)i =15000/sqrt(30600/g) = about 268m/sec
So the muzzle velocity would be 536 m/s right?

Vxi=268 m/s, that is right. But the muzzle velocity is not Vxi+Vyi. Vxi and Vyi are the horizontal and vertical components of the velocity, a vector. Its magnitude is Vm=√(Vxi2+Vyi2).

Cassomophone said:
Now I am not sure where to go on part B. All i know is the muzzle velocity is doubled so that should be [STRIKE]2(536)= 1072 m/s [/STRIKE]

If the muzzle velocity is doubled, both the horizontal and vertical components are doubled, both become 536 m/s.

Cassomophone said:
Now θ=45°
g=9.8 m/s^2
Xf=x+5

Xf=x+5000(meter)

Cassomophone said:
Xi=0
Yf=0
Yi=3km or is this 0 now since it's further back from the cliff?
Yi=300 m. The canon is still on the cliff, it has changed its horizontal position only.

Cassomophone said:
[STRIKE](Vx)f and (Vy)f=0[/STRIKE]
Δt=? (Initial time is 0)
(Vx)i and (Vf)i=536 m/s
[STRIKE]Vm= 1072 m/s[/STRIKE]

The cannon ball has to cover x+5000 horizontal distance with the horizontal velocity 536 m/s. That takes time t. 536t=x+5000. Isolate t and substitute it into the equation for y.

ehild
 

1. How does the professor's crazy cannon question relate to real-world situations?

The professor's crazy cannon question is a hypothetical scenario used to illustrate the principles of physics and how they can be applied to solve problems in the real world. It challenges students to think critically and creatively, and to apply their knowledge to practical situations.

2. What is the purpose of the crazy cannon question?

The purpose of the crazy cannon question is to engage students in active learning and to promote deeper understanding of the concepts being taught. It encourages students to think outside the box and to apply their knowledge in new and challenging ways.

3. Is the crazy cannon question solvable?

Yes, the crazy cannon question is solvable. It may seem complex and challenging at first, but by breaking it down into smaller parts and applying the principles of physics, it can be solved. It may require some trial and error and critical thinking, but there is always a solution.

4. How can I approach the crazy cannon question?

To approach the crazy cannon question, it is important to first understand the principles of physics involved, such as projectile motion and conservation of energy. Then, it is helpful to break down the problem into smaller parts and use equations and diagrams to visualize the situation. It may also be beneficial to work with a group or seek help from the professor or teaching assistant.

5. Can I use the crazy cannon question to study for exams?

Yes, the crazy cannon question can be a valuable tool for studying for exams. It challenges students to apply their knowledge and problem-solving skills, which are essential for success in physics exams. By practicing similar questions and understanding the concepts behind them, students can improve their understanding and performance on exams.

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