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Help understand

  1. Sep 16, 2005 #1
    Hi,
    Please help me to understand this question of the problem:

    A boy wants to knock down a coconut with a rock and his slingshot. He observes that the coconut is about 3.0m above his slingshot and the tree is 4.0m away along the ground.
    He knows from experience that the release speed of his rock is 20m/s.
    How far above the coconut should he aim?

    I don't understand what I am asking to do here. Please can someone help me?

    brad
     
  2. jcsd
  3. Sep 16, 2005 #2

    Doc Al

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    Staff: Mentor

    If gravity didn't exist, the boy would aim directly at the coconut, since the rock would travel in a straight line. But gravity does exist. Realize that, compared to where it would have gone with no gravity, the rock falls a certain distance. That distance is how far above the monkey you have to aim.

    Another way to look at it: What angle must the rock be shot at to hit the coconut? Once you find that angle, you can see where the straight line path would have been.
     
  4. Sep 16, 2005 #3
    Doc, I have a problem to find the angle. I try to use the equations of motion but I have too many unknow : tfinal, vy/final, and the angle itself.

    vy/final=vy/initial-g(ty/final)

    yfinal=yinitial+vy/initial*tfinal-.5*(g)yfinal2

    those two equations give respectively:

    vy/final=20*sin(θ )-9.81*tfinal

    4=0+ 20sin(θ )*tfinal-4.9*tfinal2

    Do i miss something here?
    please help

    brad
     
  5. Sep 16, 2005 #4

    Doc Al

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    You have the vertical distance as a function of time. Good! But what about the horizontal distance?
     
  6. Sep 16, 2005 #5
    xfinal=xinitial+vx/initial*tfinal
    That gives:
    3=0+20cos(θ )* tfinal
     
  7. Sep 16, 2005 #6

    Doc Al

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    Good! (I think you mean 4, not 3.) Now combine that with the equation for vertical motion.
     
  8. Sep 16, 2005 #7

    Doc Al

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    Two things:
    (1) I think you have mixed up the 3 and the 4. According to your first post:
    (2) Here's a trick that may help you combine those two equations. For each equation, isolate the term with the sin or cos. Then square both sides of each equation. Then add them. (I assume you know a useful trig identity about [itex]\sin^2\theta + \cos^2\theta[/itex].)
     
  9. Sep 16, 2005 #8
    Well I found somthing that is not realistic
    I change the 3 and 4 into the right equations and I found t=3.92s

    and the angle is .99 degree!!

    What do you think?
     
  10. Sep 16, 2005 #9

    Doc Al

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    Well... I didn't crank out the numbers myself, but does your answer make any sense? After all, without gravity the angle would be [itex]\tan \theta = 3/4[/itex]. So with gravity, the angle must even be greater. Recheck your calculations. (I'll do it myself when I get a few minutes.)
     
  11. Sep 16, 2005 #10
    ok i'll do that
     
  12. Sep 17, 2005 #11

    Doc Al

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    I did the calculation and found, as expected, that the angle is slightly greater than that needed for straightline motion. (Note: When solving the quadratic equation, there are two solutions. Only one of them is the one we want.) If you still get an unrealistic answer, post the steps just as you did them.
     
  13. Sep 17, 2005 #12
    I have te two equations:

    3=0+ 20sin(θ )*tfinal-4.9*tfinal2

    4=20*cos(θ)*tfinal

    Isolate the sin and cos

    sin(θ)=(3+4.9*tfinal2)/20*tfinal

    cos(θ)=4/20*tfinal

    I combinethe squares of sin and cos

    cos(θ)2+sin(θ)2=(9+29.4*tfinal2+24.01*tfinal2+16)/(400*tfinal2)

    equivalent to

    1=(25+29.4*tfinal2+24.01*tfinal4)/(400*tfinal2)


    equivalent

    (25+29.4*tfinal2+24.01*tfinal4)=400*tfinal2



    (25-370.6*tfinal2+24.01*tfinal4)=0

    solving this I found 4 values of t but one is correct t=3.92s

    Please tell me what is wrong here.
     
  14. Sep 17, 2005 #13

    Doc Al

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    Your work looks correct. Treat the final equation, as I'm sure you did, as a quadratic in t^2 (say X = t^2). The quadratic has two solutions: you just picked the wrong one! (When you take the square root of those solutions, you can ignore the negative values.)

    There are two ways to hit the coconut: The long way or the short way. The long way is essentially shooting it up in the air in tall arc. That's not the one you want.
     
  15. Sep 17, 2005 #14
    I tried your suggestions
    I checked and rechecked, I found the same thing.
    I rechecked the equation too. they sound fine!
    :cry:
     
  16. Sep 17, 2005 #15

    Doc Al

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    Please list those four solutions.

    Alternatively, if I view this as a quadratic in x = t^2:

    [tex]24.01x^2 - 370.6x + 25 = 0[/tex]

    This equation has two solutions. What are they?
     
  17. Sep 17, 2005 #16
    the four solutions I found are:
    x=3.9201...
    x=0.2602....
    x=-0.2602...
    x=-3.9201

    I get rid of the negative values and I have:

    x=3.9201...
    x=0.2602.... but I only take x=3.9201...

    when I use you equation , I have
    x=15.36...
    x=0.06775...
     
  18. Sep 17, 2005 #17

    Doc Al

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    And why do you ignore the other answer? That's the one you want!

    Right. And when you take the square roots, you get the same four solutions.
     
  19. Sep 17, 2005 #18

    lightgrav

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    Look, the coconut is 5m away from the start
    (along a diagonal). If the stone travels 20m in 1 sec
    then (ignoring gravity) it takes about 1/4 sec to go 5m.

    WITH gravity, in 1/4 sec the stone deviates from its path
    by 0.3 m, so you have to aim about twice as high,
    and stone will take anout twice as long ... about half sec.

    Why do you inisist on discarding the t^2 = 0.2602 [s^2]
    (which means t about .51 sec) ?
    You know that during 2 seconds of free-fall,
    a stone would deviate from its path by nearly 20 meters
    - so you want to aim 20 meters high?
    That's what DocAl meant by "the long way" almost straight up.
     
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