Projectile Motion: Solving for Optimal Cannon Angle in Avalanche Triggering

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Homework Help Overview

The discussion revolves around determining the optimal angle for firing a cannon to trigger an avalanche, given specific parameters such as muzzle speed, horizontal distance, and height difference. The subject area pertains to projectile motion in physics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formulation of an equation based on projectile motion, with attempts to derive a quadratic equation using trigonometric identities. Some express confusion over specific factors in the equation and question the correctness of their attempts.

Discussion Status

The discussion is ongoing, with participants providing hints and suggestions for alternative methods. There is a recognition of the need to verify calculations, and some participants are exploring different interpretations of the problem.

Contextual Notes

Participants mention constraints such as the inability to use graphing calculators and the pressure of a tight course schedule, which may affect their problem-solving approach. There is also a discussion about the relevance of certain variables in the context of the problem.

babysnatcher
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A cannon with a muzzle speed of 1000 m/s is used to start an avalanche on a mountain slope. The target is 2000 m from the cannon horizontally and 800 m above the cannon. At what angle, above the horizontal, should the cannon be fired?
 
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Where is your attempt at a solution. No try, no help.
 
barryj said:
Where is your attempt at a solution. No try, no help.

lol, ok.

help me solve this then :

1600(cos(x))^2=1000(4000)sin(x)cos(x)-9.80(2)^2

^ le attempt xD
 
I believe this equation is correct. To solve, you must use a trig identity to arrive at a quadratic equation to solve.
 
barryj said:
I believe this equation is correct. To solve, you must use a trig identity to arrive at a quadratic equation to solve.

how do I do that? xD

the reason why I got stuck is because of that junk.

I need help solving that or an alternative method to solve for the angle.
 
Well, if you don't want to do the trig thing, then use a graphing calculator.
 
barryj said:
Well, if you don't want to do the trig thing, then use a graphing calculator.

we( the class) can't "graphing calculator" O:
 
babysnatcher said:
1600(cos(x))^2=1000(4000)sin(x)cos(x)-9.80(2)^2

It looks almost right. Please double-check.
 
This is correct. OK, I'll give another hint. Divide everything by cos^2(x) then look for the trig identity to make a quadratic equation to solve.
 
  • #10
lewando said:
It looks almost right. Please double-check.

looks very right ^.^
 
  • #11
I'm having trouble understanding the origin of the 1000 factor:

"1600(cos(x))^2=1000(4000)sin(x)cos(x)-9.80(2)^2"

I'll check my math again in the morning--with fresher eyes.

barryj's suggestion is a good one.
 
  • #12
yeah, I don't think that 1000 should be there either
 
  • #13
The 1000 should not be there. Did you all figure out the required trig idenity?
 
  • #14
There is something wrong with the equation you gave as others says.
Btw does it says you to use 9.8 for g? If not use 10 instead of 9.8
 
  • #15
I was taught this weeks ago in my college.
Teacher got the answer by derivation.
Last answer was tan@ = u / (u^2 - 2gh)^1/2
Where u is 1000
g is gravity
h is 800
 
  • #16
Consider that sec^2(x) = 1 + tan^2(x)
 
  • #17
Bandarigoda said:
I was taught this weeks ago in my college.
Teacher got the answer by derivation.
Last answer was tan@ = u / (u^2 - 2gh)^1/2
Where u is 1000
g is gravity
h is 800

wrong answer ^.^
 
  • #18
Post 16 is the clue!
 
  • #19
Bandarigoda said:
I was taught this weeks ago in my college.
Teacher got the answer by derivation.
Last answer was tan@ = u / (u^2 - 2gh)^1/2
Where u is 1000
g is gravity
h is 800

Unfortunately, this answer does not contain a variable for horizontal distance (2000m). So...it's not helping. Even if you provided the correct answer, it probably would not help the OP who is currently trying to understand how to determine the answer, with the help of the powerful clues provided by barryj.
 
  • #20
lewando said:
I'm having trouble understanding the origin of the 1000 factor:

"1600(cos(x))^2=1000(4000)sin(x)cos(x)-9.80(2)^2"

I'll check my math again in the morning--with fresher eyes.

barryj's suggestion is a good one.

oopsie

400(cos(x))^2=1000sin(x)cos(x)-9.80

xD
 
  • #21
barryj said:
Post 16 is the clue!

i got it. can i mark this thing as solved or something like that? O:
 
  • #22
What did you get for the answer?
 
  • #23
barryj said:
What did you get for the answer?

89.43624626 and 22.36516323
 
  • #24
I checked the 22.36 answer. Did you happen to check the 89.436 answer?
 
  • #25
barryj said:
I checked the 22.36 answer. Did you happen to check the 89.436 answer?

nope, book says they are right. how should i check?
 
  • #26
babysnatcher said:
oopsie

400(cos(x))^2=1000sin(x)cos(x)-9.80

xD
That's right. (Or at least, it is the same as I got). Effectively, it is removing the factor 1000 from your old answer, and then dividing everything by 4.
 
  • #27
babysnatcher: calculate the Vx and Vy components of the 1000 m/s velocity at both angles. Then find the time to get to 2000 meters horizontally then plug this time into the S = vt + 1/2at^2 to fine the heighth at the time when it has traveled 2000 meters. In bothj cases the height should be 800 meters. I was just wondering if the 89 degree answer is an extraneous solution or is it real.

As you know, you should always try to find a way to verify your answers.
 
  • #28
barryj said:
babysnatcher: calculate the Vx and Vy components of the 1000 m/s velocity at both angles. Then find the time to get to 2000 meters horizontally then plug this time into the S = vt + 1/2at^2 to fine the heighth at the time when it has traveled 2000 meters. In bothj cases the height should be 800 meters. I was just wondering if the 89 degree answer is an extraneous solution or is it real.

As you know, you should always try to find a way to verify your answers.

ewww, why would i always have to do that. this course is 6 weeks, I am already doing extra problems so I am behind. no time xD
 
  • #29
haha. How much time is there in a day? Probably enough. If you get into the habit of checking your answers, then the more you practice it, the quicker you will be able check them. And the quicker you will be at doing new problems, so I don't think it is a waste of time.
 
  • #30
BruceW said:
haha. How much time is there in a day? Probably enough. If you get into the habit of checking your answers, then the more you practice it, the quicker you will be able check them. And the quicker you will be at doing new problems, so I don't think it is a waste of time.

wake up, get ready - 2 hours

sleep - 8-10 hours

break time - 4 hours

bus rides - 2 hours

daily( M-F) class - 3 hours

so about 5 hours xD.

physics chapters - 13, and I am on chapter 4 right now and its week 3. o.O she is barely lecturing on chapter 5/6.
 

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