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Wrong analysis in projectile problem

  1. Feb 13, 2012 #1
    An archerfish shoots at a target 2.00 m away, at an angle of 30.0o above the horizontal. With what velocity must the water stream be launched if it is not to drop more than 3.00 cm vertically on its path to the target?


    i have troubles with this problem (Physics for scientists and engineers 6th edition)(Serway, Jewett)
    i need to know what is wrong in my analysis



    My Answer
    target position x=2cos30=1.73 m. , y=2sin30=1 m.
    maximum height H=1+0.03=1.03 m.
    but H=vi2sin2θi/2g
    => vi2sin2θi=(9.8)(2.06)
    the equation of the trajectory is y=(tanθi)x-[g/(2vi2cos2θi)]x2
    => 1=1.73tanθi-[(9.8)(1.5)/(vi2cos2θi)]
    substituting for the value of vi2
    we get a quadratic equation in tanθ
    0.73tan2θi-1.73tanθi+1=0
    either θi=54° or θi=45°
    hence vi=5.6 m/s or vi=6.35 m/s

    this solution is wrong , i don't know why?
    the solution manual answer is 25.8 m/s with a different approach in the problem solving
    so what is wrong in my solution
     
  2. jcsd
  3. Feb 13, 2012 #2
    Shouldn't that Max Height be 1-0.03 meters = 0.97 meters?
     
  4. Feb 13, 2012 #3
    of course not, if the maximum height is 0.97 this means that the water shoot will never get the target which is at height of 1 m.
    even trying to substitute for H=0.97 will result in imaginary values for tanθ
     
  5. Feb 13, 2012 #4
    you're given an angle of 30 degrees and told that it must not drop more than 3 cm...sounds like 0.97 meters since the question you quoted stated "drop", not rise...but it's your karma...good luck
     
  6. Feb 13, 2012 #5
    Hi TxRationalist
    i think that you agree with me that the target is at height of 1 m.
    and because the problem statement saying that the water drops 3 cm before catching the target, and since the projectile rising then dropping so the maximum height must be larger than the target height , (water falls from the point of maximum height to a distance 3 cm before catching the target)
    so the maximum height=target height + falling distance=1+0.03=1.03

    more than that i tried your answer for the height to be 0.97 m. and it yeilds imaginary values for tanθ

    anyway thanks for your help :)
     
  7. Feb 13, 2012 #6
    I'm not saying I'm right, just that that is what jumped out at me as the path to the solution when I looked at it. I ran the numbers and the quadratic equation gave me a negative for "b squared minus 4ac" also. Unless I did something wrong. i'm going to take another look at it.
     
  8. Feb 13, 2012 #7
    well, i didn't cancel out the "t" the first time, so that my quadratic equation was in the general form. I corrected that, which led to the form of at^2+c=0. I solved this using my 0.97 m and got a result of 25.7 m/s. Awfully close. I'm going to run it with the same method using your numbers. Just so you know my method, I used the x component of the velocity to solve for V0, substituted this into my Vy equation and solved that t. I then substituted this back into the x/[t*cos(theta)] to get 25.7 m/s
     
  9. Feb 13, 2012 #8
    sorry i didn't understood what you did, if you please help me with a complete solution, let me know what is wrong in my solution
    many thanks
     
  10. Feb 13, 2012 #9
    I broke the problem down into the x and y components:

    x = (V0cosθ)t
    y = (V0sinθ)t - 0.5gt^2

    I used the first equation to isolate V0, substituted this into the second equation and solved for t. I then used the first equation again to get my answer for V0.

    Hope this helps, I have to get back to work.
     
  11. Feb 13, 2012 #10
    may i am mistaken, but i wonder how you solved two equations for three variables, V0,θ and t
     
  12. Feb 13, 2012 #11
    what is wrong in my solution?, this is what makes me go mad
     
  13. Feb 14, 2012 #12
    theta = 30 degrees.
     
  14. Feb 14, 2012 #13
    Dear TxRationalist
    theta that you are mentioning in your equation is not 30°(30° which is the angle the position vector of the target with respect to the fish makes with the horizontal) , in your equations it must be θi not θ, and θi is not given,
    even the solution manual saying the same thing
    and it calculated θi using a different approach θi=30.7°

    anyway thanks for your kindness and your tries to help
     
  15. Feb 14, 2012 #14

    vela

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    Is the wording of the problem exactly as it was given in the book? It's a bit ambiguous, and I wouldn't interpret it the way you have. Was there an accompanying figure you could provide?
     
  16. Feb 14, 2012 #15
    unfortunately there is no accompanying figure , and i agree with you that the wording of the problem is ambiguous

    anyway here is the complete text of the problem

    The small archerfish (length 20 to 25 cm) lives in brackish waters of southeast Asia from India to the Philippines. This aptly named creature captures its prey by shooting a stream of water drops at an insect, either flying or at rest. The bug falls into the water and the fish gobbles it up. The archer-fish has high accuracy at distances of 1.2 m to 1.5m, and it sometimes makes hits at distances up to 3.5 m. A groove in the roof of its mouth, along with a curled tongue, forms a tube that enables the fish to impart high velocity to the water in its mouth when it suddenly closes its gill flaps. Suppose the archerfish shoots at a target 2.00 m away, at an angle of 30.0° above the horizontal. With what velocity must the water stream be launched if it is not to drop more than 3.00 cm vertically on its path to the target?
     
  17. Feb 14, 2012 #16

    vela

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    OK, I think you're interpreting it correctly, and the target is at ##x=(2.00\cos 30^\circ)\text{ m}## and ##y=(2.00 \sin 30^\circ)\text{ m}##.

    The mistake you're making is assuming that the maximum height the stream achieves is 1.03 m.

    The equation for y(t) is ##y(t) = v_{0_y} t - \frac{1}{2}gt^2##. If there were no gravity, the height of a piece of water would be equal to ##v_{0_y} t##. Because there is gravity, it's actually at a height ##v_{0_y} t - \frac{1}{2}gt^2##. The difference between those two, ##\frac{1}{2}gt^2##, is the distance it has fallen. That's what you want to set equal to 3.00 cm.

    Note that the water could still be on its way up when it has fallen 3.00 cm. That's why you can't assume the water reaches a height of 1.03 m.
     
  18. Feb 14, 2012 #17
    how come!!!?

    me too think that i have troubles with the maximum height , but the wording of the problem let me think that fallen means going from the point of maximum height to the point where the target exist
     
  19. Feb 14, 2012 #18

    cepheid

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    What integral is saying is that the amount the projectile has "dropped" is the difference in height between the linear trajectory that the projectile would take if there had been no gravity, and the parabolic trajectory that it instead takes in the presence of gravity.

    So as soon as it is launched, it begins to "drop" relative to this straight path.

    See the attached figure

    http://img542.imageshack.us/img542/9411/dropp.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  20. Feb 14, 2012 #19
    aha
    i understand now what "vela" and "cepheid" mean
    and this means that i can't make any assumption about the maximum height, water may hit the target before the maximum height or after the maximum height, i think what confused me is that the word "drop" which seems to me very confusing (i should understand that it is a drop from the straight line path without gravity)
    now i can get θi from the larger triangle it θ=tan-1 (1.03/1.73)=30.7°
    now i can solve the equations for the time t and the initial velocity vi

    its clear now :)

    many thanks guys, i think i was't smart enough to interpret the problem statement
     
  21. Feb 14, 2012 #20
    By the way the authors of the book avoided this problem in 7th and 8th editions of the book
     
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