Projectile Motion: Solving for Optimal Cannon Angle in Avalanche Triggering

[babysnatcher]

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?

 

[barryj]

Where is your attempt at a solution. No try, no help.

 

[babysnatcher]

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

 

[barryj]

I believe this equation is correct. To solve, you must use a trig identity to arrive at a quadratic equation to solve.

 

[babysnatcher]

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.

 

[barryj]

Well, if you don't want to do the trig thing, then use a graphing calculator.

 

[babysnatcher]

Well, if you don't want to do the trig thing, then use a graphing calculator.

we( the class) can't "graphing calculator" O:

 

[lewando]

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

It looks almost right. Please double-check.

 

[barryj]

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.

 

[babysnatcher]

It looks almost right. Please double-check.

looks very right ^.^

 

[lewando]

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.

 

[BruceW]

yeah, I don't think that 1000 should be there either

 

[barryj]

The 1000 should not be there. Did you all figure out the required trig idenity?

 

[Bandarigoda]

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

 

[Bandarigoda]

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

 

[barryj]

Consider that sec^2(x) = 1 + tan^2(x)

 

[babysnatcher]

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 ^.^

 

[barryj]

Post 16 is the clue!

 

[lewando]

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.

 

[babysnatcher]

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

 

[babysnatcher]

Post 16 is the clue!

i got it. can i mark this thing as solved or something like that? O:

 

[barryj]

What did you get for the answer?

 

[babysnatcher]

What did you get for the answer?

89.43624626 and 22.36516323

 

[barryj]

I checked the 22.36 answer. Did you happen to check the 89.436 answer?

 

[babysnatcher]

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?

 

[BruceW]

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.

 

[barryj]

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.

 

[babysnatcher]

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

 

[BruceW]

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.

 

[babysnatcher]

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. she is barely lecturing on chapter 5/6.

 

[barryj]

Hee is how to do it..

Wake up and get ready in 1 hour
reduce sleep to 7 hours max
work problems on the bus

Now you have extra time to check answers.

Seriously, you should get into the habit of checking answers for all assignments, not just physics. When you finally get a real job, I assume you will, your boss will want correct answers, not an attempted solution.

BTW what level course are you taking? HS, college, or what?

 

[babysnatcher]

Hee is how to do it..

Wake up and get ready in 1 hour
reduce sleep to 7 hours max
work problems on the bus

Now you have extra time to check answers.

Seriously, you should get into the habit of checking answers for all assignments, not just physics. When you finally get a real job, I assume you will, your boss will want correct answers, not an attempted solution.

BTW what level course are you taking? HS, college, or what?

nah, need that sleep or my brain doesn't function properly and studying takes longer.

Physics 1 - Mechanics of Solids

hmm, when should i give up on a problem and ask for help? after how much time? i just go by how easy it looks. this problem didnt seem this difficult so i kept trying for 3 hours.

 

[barryj]

Is mechanics of soloids mostly Newton laws of motion? Is this a college course?" Where?

 

[babysnatcher]

Is mechanics of soloids mostly Newton laws of motion? Is this a college course?" Where?

im going to name the chapters except for the first 4.

the laws of motion
circular motion and other applications of Newton's laws
work and kinetic energy
potential energy and conservation of energy
linear momentum and collisions
rotation of a rigid object about a fixed axis
rolling motion and angular momentum
static equilibrium and elasticity
fluid mechanics

she is skipping oscillatory motion and the law of gravity because its a 7 week course and this is a semester school, i guess. its suppose to be 15 chapters for physics 1.

East los angeles college.

k, so what variable do i solve for with the angle? do i just arbitrarily pick v, x, or y to check my answer?

 

[BruceW]

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. she is barely lecturing on chapter 5/6.

mm. It is pretty packed when you write it like that. My advice - keep in mind that you have 5 hours a day for out-of-class study, and make sure that you have done the important stuff within this time (for example, doing the set problems), and if you have some of that 5 hours left over, then you can do the less important stuff (like reading other physics texts, and doing extra problems).

I knew a guy who worked 17 hours per day (and he did the same amount on weekends too). He was manager of a pub, and he had to be working at the pub literally any time that the pub was open. Pretty crazy.