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Finding two variables in projectile motion

  1. Sep 19, 2015 #1
    1. The problem statement, all variables and given/known data
    You are the chief of the Angry Birds (Google it, if you don’t know the reference...it’s not important). You can shoot red birds from a catapult, and they will fly as projectiles under the effect of gravity. Gravity points downwards and has magnitude g = 9.80 m/s2 . Your goal is to hit some nasty green pigs, who have stolen your Angry Bird Eggs. You are able to adjust the angle θ and speed v of the projectile, as it leaves the catapult. The projectile leaves the catapult at a height h above the ground.

    proxy.php?image=https%3A%2F%2Fi.gyazo.com%2Fb855d7c86b55ca41dfb05406fd07a683.png

    a) First, consider the case when a single green pig is placed on the ground, at a distance d from the foot of the catapult. Find a relation between the angle and speed required to hit it. Remember to draw a sketch of the situation.

    b) Next consider the case where there are two green pigs. One on the ground at a distance d as before, the other at a distance l < d, and placed on a pole of height h. What should you choose for θ and v in order to hit both pigs 1 with the same red bird? Why must one require l < d? Remember to draw a sketch of the situation.

    2. Relevant equations
    This task is related to projectile motion and I'm therefor pretty sure I'm supposed to solve it with the projectile motion equations.
    In other words x = Vx(initial)*t because there is no acceleration in the x-axis direction and y = Vy(initial)t - 1/2gt^2
    This is by the way a question i got at university but I'm posting it here because the physics seen rather basic, the problem is rather that in task b i need to determine two variabels and I'm not given the distance. My point is that if calculus or som other more advanced methods like advanced algebra could be used to solve this then that should not be a problem for me.
    g = gravity = 9.8 m/s^2
    d is the distance between 0 and the last pig
    θ = is the angle the projectile was shot out with

    3. The attempt at a solution
    In task a I found an expression for t in x-direction by using the x = Vxt (and setting x to d) and got t = d/Vx
    I put this expression for t in to the equation in y-dirextion: y = Vyt - 1/2gt^2. By setting y to -h because i think like the projectile is shot from h which is 0 and land on the ground, -h. Also I've used that Vx = Vcosθ an Vy = Vsinθ
    I solved this equation for V and as result I got:
    V = sqrt((g*d^2)/(2*cos^2(θ)(h+d*tanθ)))
    This is the answer I think is correct in task a and I think I might have to use it to solve b but im not sure.
    The b task is difficult because I need to find both θ and the innitial speed in order to hit both pigs an by my logic there is an endles different possible answers to that.
    I've been trying to find equations for t or v but I have no idea how to make sure it hits both targets and write an equation on that so if anyone think they can solve it or help me by giving me som hints or pointers to where i should start at least that would be most appreciated. Also sorry for my bad english but I hope you understand
     
  2. jcsd
  3. Sep 19, 2015 #2

    TSny

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    Your result for part (a) looks good to me.

    For part (b), try deriving a similar relation between V and θ for hitting the bird on the pole. Then you will have two equations for the two unknowns V and θ.
     
  4. Sep 19, 2015 #3
    I tried to do the same thing as in a but replacing d with l instead and setting h to 0 because the projectile is shot and is landing in the same height so therefor I put it to 0
    Got this equation: 0=l*tanθ - g/2*(l^2)/(v^2cos^2(θ)). Should I also solve this equation for v you think or how should I find v and θ? θ seems especially difficult to me because theres both tanθ and cos^2(θ)
     
  5. Sep 19, 2015 #4

    TSny

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    Looks good. You might try rearranging this to get g/2*(l^2)/(v^2cos^2(θ)) on one side.
    Similarly, from your relation in part (a), solve for g/2*(d^2)/(v^2cos^2(θ)) .

    See what happens if you then divide the two equations.
     
  6. Sep 20, 2015 #5
    585405a5f0a69e44cecc9b549d150f03.png
    d58fa6f89188170060985d6dd6b65e4b.png
    This is the two equations I then get. I've never devided two equations on eachother so not quite sure how to do it but depending on whitch of them i devide on the other I will either get l/d + h or d/l + h on the right side. The left I don't know, it seems to me I get l^2/d^2 or the other way around. Was this what you ment by divide the two equations?
     
  7. Sep 20, 2015 #6
    You can put tanθ/l=tanθ/d+h/d^2 and solve for θ
     
    Last edited: Sep 20, 2015
  8. Sep 20, 2015 #7

    TSny

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    Yes. Or you can divide the top equation by ##l^2## and the bottom equation by ##d^2##. Then the left hand sides are equal.
     
  9. Sep 22, 2015 #8
    Is the answer θ=tan^-1(-hl/ld-d^2) ?
     
  10. Sep 22, 2015 #9

    TSny

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    EDIT: Well, I don't want to say if this is the correct answer because Makonia has not yet posted an answer.

    EDIT 2: Now that Makonia has given her answer, I can comment that I think your answer is correct. By the way, welcome to PF!
     
    Last edited: Sep 23, 2015
  11. Sep 22, 2015 #10
    I got the same thing. But not sure if it is correct, nor the best answer.
     
  12. Sep 22, 2015 #11
    I got θ=tan^-1(lh/d(d-l)) so I think it's correct
     
  13. Sep 22, 2015 #12

    TSny

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    OK. Great. I think that's the answer, too.
     
  14. Sep 23, 2015 #13
    Yes, but then the right sides will be: tan(θ) / l) i the first eq. and in the second: (tan(θ) + h ) / d, and when you put these two equal to one another the tan will go away?
    So how did you then get θ=tan^-1(lh/d(d-l)) ?
     
  15. Sep 23, 2015 #14

    TSny

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    The tan will not go away due to the fact that tan on one side is divided by l and tan on the other side is divided by d.
     
  16. Sep 23, 2015 #15
    Hi i'm only getting tan^-1( lh/d(1-l ), i don't understand how you guys find "d^2'" in the nominator, please help...
     
    Last edited: Sep 23, 2015
  17. Sep 23, 2015 #16

    TSny

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    Welcome to PF!

    I believe you are saying that your denominator is d - dl. But note that it doesn't make sense to subtract these two terms. d had the dimensions of length while dl has the dimensions of length squared. If you show us your individual steps, we can track down your source of error.

    As a sort of nit-picky point, parentheses should be used around the denominator as so: tan^-1( lh/(d-dl )). Otherwise, it would be interpreted as dividing lh by just d rather than (d - dl). Also, there are formatting icons for superscripts, so you can write it as tan-1[ lh/(d-dl )].
     
  18. Sep 23, 2015 #17

    TSny

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    Welcome to PF! It's good to see that we are all getting the same result.
     
  19. Sep 23, 2015 #18
    I'm sorry I wrote my answer in the stress and chaos of all this. My answer is of course tan^-1( lh/d(1-l )) (NOT "d-dl" in denominator). This will be length vs. length, so the dimensions adds up. But if i put in imaginary numbers for l, h & d. I'm getting a negativ angel, so my calculation must be wrong.
    I'll show you how I have done it:
    tanΘ/l = tanΘ+h/d
    I multiply with l
    and move l*tanΘ to the other side.
    tanΘ-l*tanΘ=lh/d
    tanΘ(1-l)=lh/d
    Tan^-1(lh/d(1-l))
     
  20. Sep 23, 2015 #19
    You have tanΘ/l = tanΘ/d+h/d^2
    Then you end up with tan^-1(lh/d(d-l))=Θ
     
  21. Sep 23, 2015 #20
    oooh that's why, thnx man :-D
     
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