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Finding Angles in Projectile Motion

  • Thread starter rakeru
  • Start date
  • #1
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Homework Statement



A fire hose held near the ground shoots water at a speed of 6.8 m/s. At what angle(s) should the nozzle point in order that the water land 2.0 m away? Why are there two different angles?

ax=0
xo=0
x=2.0 m

ay= 9.80 m/s/s
y0=0
y= I'm guessing 0, because it lands on the ground at approximately the same level..



Homework Equations



I'm guessing:

x = Vx0t

Vx0 = 6.8 cosθ

Vyo= 6.8 sinθ

y= y0 + Vyot - 1/2ayt2



The Attempt at a Solution



Oh my god. I have no idea why I'm having so much trouble with this problem. I've been trying to solve this for about two hours. I've re-written the equations in my book multiple times and I don't know how to solve this. I've been trying to solve for the time but there are so many variables.. I'm not even really familiar with cosine and sine. The problem implies that there are two angles in the answer. So I'm guessing it's because there's some sort of quadratic equation involved. I don't know. :/

I even tried to find a way to use substitution and find the time, but it was no use. Please, help! :'(
 

Answers and Replies

  • #2
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You can use the last equation to find the time (as function of θ, everything else is known), and find x based on that. As y and y0 are 0, you do not have to solve a (real) quadratic equation.
 
  • #3
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You can use the last equation to find the time (as function of θ, everything else is known), and find x based on that. As y and y0 are 0, you do not have to solve a (real) quadratic equation.
I don't understand.. how would I do that?? I already know that x is 2.0m.. Right? The final position is 2.0 meters away.. If I don't know Vyo, then how can I find t?? I'm so confused. Maybe it's the math :/
 
  • #4
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9,660
how would I do that??
Replace Vx0 and t by their formulas - you have the one for Vx0, you just need one for t.
I already know that x is 2.0m
Exactly, and you can use this to solve for the angle afterwards.

Anyway, try to use the last equation first. Plug in everything you know about the parameters there, simplify, try to solve for t. If you get stuck somewhere, post your steps here please.
It is just mathematics, indeed.
 
  • #5
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When I plug everything in, should I have:

0=6.8sinθ - 1/2(-9.80)(2.0/6.8cosθ)2

? Because t= x/Vx0
 
  • #6
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Uhm, so I did this:

0 = 6.8sinθ + 4.90[4.0/(6.8cosθ)2]

=6.8sinθ+ 19.6/(6.8cosθ)2

=(6.8sinθ)(6.8cosθ)2+ 19.6/(6.8cosθ)2

But I don't know what to do now.. Do I just remove the 6.8cosθ2?
 
  • #7
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9,660
Don't replace t in the last equation, that makes it more complicated.

0 = Vyot - 1/2ayt2
Can you find solutions for t in this equation? Something like t=...?
One solution is given by t=0 - it has to be, as you start at the ground. But there is another solution.
 
  • #8
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So like:

4.9t^2 + 6.8sin(theta)t = 0

t(4.9t + 6.8sintheta)= 0

t=0 or t= -6.8sin(theta)/4.9

?? Thank you so much for helping me, by the way. If what I did is right, then do I put that into 2.0= Vx0t??
 
  • #9
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9,660
If what I did is right, then do I put that into 2.0= Vx0t??
Right. This will give you an equation where just θ is unknown.
 
  • #10
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Okay but now the problem is that I don't know what to do with the cos and sin.. Is there a number for each? Is sin theta just multiplying the angle by something? How would I even multiply cos and sin together ? :/
 
  • #11
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9,660
Is sin theta just multiplying the angle by something?
It is not.

It could be enough to give that equation a computer (WolframAlpha can solve this, for example).
If you have to show more steps, you can use ##\sin^2(\theta)+\cos^2(\theta)=1## to get rid of either sin or cos, and solve the equation for sin(θ) (or cos(θ)) afterwards.
 
  • #12
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Well, when subbing your value for t into the equation for the displacement in the x direction, isn't there a trig identity that'd be much simpler to use?
 
  • #13
verty
Homework Helper
2,164
198
Well, when subbing your value for t into the equation for the displacement in the x direction, isn't there a trig identity that'd be much simpler to use?
There is yes, one of the rules you need to know. It is actually in reverse.
 
  • #14
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There is yes, one of the rules you need to know. It is actually in reverse.

But it is time for you to work it out now. Not much work has been shown in this thread.
I'm not actually the one asking the question. I was just trying to make the asker aware that there's a relatively simple solution available.
 
  • #15
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Okay so from:

9.8 = (6.8cosθ)(-6.8sinθ)

How would I put in

cos^2θ = 1 - sin^2θ ?

If I have cosθ, how would I make that cos^2θ to put it in? I'm confused.. I'm in a college algebra class and it's basically one of the pre-requisites in math for graduating. It's the lowest college level math class and we are doing mostly functions. I don't know pre-calculus or anything. The class just started a week ago.. so yeah. :/ My physics class has no pre-requisite.. except intermediate algebra.
 
  • #16
cepheid
Staff Emeritus
Science Advisor
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Okay so from:

9.8 = (6.8cosθ)(-6.8sinθ)

How would I put in

cos^2θ = 1 - sin^2θ ?

If I have cosθ, how would I make that cos^2θ to put it in? I'm confused..
Keep in mind that cos2θ is just a shorthand. It means (cosθ)2. Therefore$$\cos\theta = \sqrt{\cos^2\theta}$$

I'm in a college algebra class and it's basically one of the pre-requisites in math for graduating. It's the lowest college level math class and we are doing mostly functions.
Well then you're in luck, because sine and cosine are functions! Trigonometric functions. It sounds like you need to brush up on trigonometry. I highly recommend doing some reading about sine and cosine online before proceeding further with this type of problem.
 
  • #17
310
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Okay so from:

9.8 = (6.8cosθ)(-6.8sinθ)

How would I put in

cos^2θ = 1 - sin^2θ ?

If I have cosθ, how would I make that cos^2θ to put it in? I'm confused.. I'm in a college algebra class and it's basically one of the pre-requisites in math for graduating. It's the lowest college level math class and we are doing mostly functions. I don't know pre-calculus or anything. The class just started a week ago.. so yeah. :/ My physics class has no pre-requisite.. except intermediate algebra.
Well, there are quite a few identities out there that can come in handy when solving variables which are part of trigonometric functions. Have a look around and I'm sure you'll find one that'll help you out. It's one of the more common ones.
 
  • #18
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I think I might just give up on this problem. It's just making me anxious. I can't really "brush up" on trigonometry... I've never really learned it.
 
  • #19
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I think I might just give up on this problem. It's just making me anxious. I can't really "brush up" on trigonometry... I've never really learned it.
Have a think about the identity sin(2x) = 2sinxcosx.
 
  • #20
verty
Homework Helper
2,164
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I would learn Trig, there are plenty resources available.

PS. ##sin^2 + cos^2 = 1## was never the right formula to use. Why deal with those horrid squares and square roots?
 
  • #21
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Hi! I didn't know if it would be right to reply here after a while.. But I'm so happy because today in class the professor explained this!! :D

He asked the class if they wanted him to explain it and everyone became so silent.. But I needed to know! So I'm like "YES!!!!!" and the girl in back of me agreed a few seconds after.. So he did this problem where he taught us how to make sense of the cosine and sine things to find the two angles! I am so happy.
 

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