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Grade 11 Question: finding launch angle

  1. Oct 12, 2012 #1
    The Question:

    A water balloon is fired 34 m/s from a water cannon, which is aimed at an angle of 18° above the ground. The centre of the cannon's target (which has a radius of 1.0m) is painted on the asphalt 42m away from the water cannon.
    a) Will the balloon hit the target? Justify your response with calculations that indicate where the water balloon will land.
    b) make one suggestion about how to adjust the water cannon so that the water balloon will hit the target. Justify your choice.

    For part a)

    I found total time to be 2.1 s
    Water balloon lands at 67.8 m (therefore it does not hit the target that is 42 m away)

    For part b)

    I understand that when angles are higher than 45 degrees, horizontal displacement decreases and the max vertical height increases.

    I dont know how to find the new launch angle that will allow the balloon to hit the target at 42 m. Any help is appreciated!
     
  2. jcsd
  3. Oct 12, 2012 #2

    jedishrfu

    Staff: Mentor

    show some work first. It seems you probably already the basic equations.

    Ask yourself what happens if I use 19 degrees or 17 degrees? how far from the target will it be?
     
  4. Oct 12, 2012 #3

    Okay I have:

    Vvertical = 34.0 m/s x sin18 = 10.5 m/s
    Vhorizontal = 34 m/s x cos18 = 32.33 m/s

    Dvertical= v1xt + 1/2a x t^2

    0 = (10.5 m/s x t - 4.9 m/s^2 x t ^2)

    t = 2.1 seconds

    Dhorizontal = Vh x t = 32.33 m/s x 2.1 s = 67.8 m

    Therefore distance travelled is 67.8 m, it will miss the target which is at 42 m from the cannon by 25.8 m.

    For Part B: I have no friggin clue cause u need a Launch |Angle in order to change the initial velocity ...... any ideas?
     
  5. Oct 12, 2012 #4
    Try writing [itex]D[/itex] as a function of [itex]V[/itex] and [itex]\theta[/itex]. From there, it should be easy for you to figure out how you can change [itex]V[/itex] or [itex]\theta[/itex] (or both) to get the desired [itex]D[/itex].

    You've run into the biggest mistake that intro physics students make when solving problems. You plugged in values to find variables that you are not interested in. Did you really need to know the numeric value of [itex]V_y[/itex] or [itex]V_x[/itex] to find [itex]D[/itex]?
    Solve everything symbolically and leave all the number crunching until you need a final answer.
     
  6. Oct 12, 2012 #5
    Okay I see, can u give me an example?
     
  7. Oct 12, 2012 #6
    $$x=v_xt$$
    $$y=-\frac{1}{2}gt^2 +v_{0y}t$$
    so
    $$t=0$$
    or
    $$t=\frac{2v_{0y}}{g}$$
    You also know [itex]v_x[/itex] and [itex]v_y[/itex] as a function of [itex]v[/itex] and [itex]\theta[/itex]. Combine all those together to get one equation that looks like:
    $$x=f(v,\theta)$$
    where [itex]f(v,\theta)[/itex] is a function of [itex]v[/itex] and [itex]\theta[/itex].
     
  8. Oct 12, 2012 #7
    Okay. But what i dont understand is, you are looking for a new angle u are only given the distance. Time and initial velocity change. So using your equation I cant see how to work it out to find the new angle given initial velocity for the older 18 degree angle and the new distance of 42 m.
     
  9. Oct 12, 2012 #8
    Before we go any further, please solve for x. I want to make sure you have done the math correctly.
     
  10. Oct 12, 2012 #9

    x= Vxt

    Vx (32.33 horizontal velocity for 18 degree angle) x t (2.1 s)

    x = 67.89 m
     
  11. Oct 12, 2012 #10
    You're still making the mistake of plugging in numbers.
    Pretend no numbers were given to you. Solve with letters to get x.
     
  12. Oct 12, 2012 #11
    x = Vxt

    so x = velocity x time

    x = a x b

    I still dont get what you are trying to show me
     
    Last edited: Oct 12, 2012
  13. Oct 12, 2012 #12
    If Anyone could just tell me how to do part b with the numbers I have in the problem: it would be greatly appreciated! I really need to get this question done.
     
  14. Oct 12, 2012 #13
    $$x=v_{0x}t$$
    You weren't given t or [itex]v_{0x}[/itex], so you have to find those with what you were given.
    You know
    $$y=v_{0y}t-\frac{1}{2}gt^2$$
    so you can find solve for t and get an expression involving [itex]v_{0y}[/itex] and [itex]v_{0x}[/itex].
    You also know
    $$v_{0x}=v\ cos(\theta)$$
    $$v_{0y}=v\ sin(\theta)$$
    so we plug those in for [itex]v_{0x}[/itex] and [itex]v_{0y}[/itex]. You will now have an expression for x with only [itex]v_0,\theta,g[/itex]

    Then, and only then, should you start plugging in numbers.
     
  15. Oct 12, 2012 #14

    Okay awesome. Can you should me what the final answer is?
     
  16. Oct 12, 2012 #15
    You are doing more work than necessary. Have you learned the range equation yet? This can be used when the initial and final elevation are the same.


    R = (v02/g)*sin2θ

    So if you want your range to be 42..

    42 = (342/9.8)*sin2θ

    sin2θ=(42*9.8/342)
    2θ = sin-1(42*9.8/342)
    θ = 20.86/2 = 10.43

    So the angle could either be around 10 or around 90-10.
    _________________________________________________________________________
    Garrett Stauber​
     
    Last edited: Oct 12, 2012
  17. Oct 13, 2012 #16
    I was assuming that since he hadn't used it, he doesn't know it. I was going based off what I know he knew.

    I also don't think it's too much work. Too often, intro physics classes don't emphasize enough that physics isn't simply plugging in numbers and that algebra/trig/calculus is required as well. While being a TA for non-major physics classes, students would simply see an equation and try to plug in numbers, and then have no clue why it's not working. 5 out of the 6 posts JackandJones made in this thread involve him either plugging the numbers right in to the equations or just asking for answers. I want him to learn physics, and how to do physics, not how to plug numbers into equations from physics.
     
  18. Oct 13, 2012 #17
    My post was not directed at you frog. I meant he was doing to much work by plugging in numbers too soon. Keep in mind that some people learn better from taking things apart rather than putting them together. Also, would you mind looking at my post titled "Spherically symmetric charge density given electric potential" and letting me know if you have any pointers?
     
  19. Oct 25, 2012 #18
    No no no no. I understood what was being said: I just had to incorporate it into previous equations which I found challenging: Overall I figured it out and got this: which was correct on my exam.

    Ex: Δt = 42 m /34 m/s x cosθ

    0 = v1 x sin x (42 m /34 m/s x cosθ) + ½ a(42 m /34 m/s x cosθ)
    0 = 32 x sinθ x cosθ – 4.9 x 42/34 x cos^2θ
    34^2 x sinθ = 4.9 x 42 x cosθ

    Sinθ/Cosθ = Tanθ
    Tanθ = 4.9 m/s^2 x 42 m / (34)^2
    Tan ^-1 = 10.1

    90 degrees – 10.1 degrees = 79.9 degrees

    Δdh = v1^2xsin2θ/a
    Δdh = (34)^2 x 2sin(79.9) x cos(80) / 9.8 m/s^2
    Δdh = 41 m
     
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