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Hunter and Monkey - Kinematics algebra question

  1. Nov 7, 2013 #1
    The problem statement, all variables and given/known data[/B]
    Hey guys, so here is my question:

    A huntsman fires a gun pointing at a monkey hanging on the branch of a tree. The huntsman and the base of the tree are on a flat horizontal. At the moment the huntsman fires the gun, the monkey will release it's grip on the branch of the tree. Use the initial bullet velocity Vo, horizontal distance Dx, monkey height Dy, and inital trajectory angle above.

    a)Show the monkey shouldn't have dropped

    b)For what minimum bullet velocity? Vo = (Dx/cos0)(sqrt(g/2Dy))

    The attempt at a solution
    [/B]

    I know that the bullet and monkey will meet when the vertical speed of the bullet x t + 1/2at^2 is equal to h - 1/2at^2.

    I also know the height of the monkey form the ground is VyoT = H

    I'm now sure how to do the question only working with variables,, if I had any sort of values I could do this easily. Thank you so much guys for taking the time to work with me :).
     
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  3. Nov 7, 2013 #2

    cepheid

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    Huh? The bullet and monkey will meet when their x and y positions are the same. Vertical speed has nothing to do with it, and in any case, the equations you posted above aren't for vertical speed.

    Figure out the equations for the x position of the bullet vs. time, the y position of the bullet vs. time, and the y position of the monkey vs. time.

    No, this is wrong. The monkey is not falling at a constant speed. It is accelerating due to gravity.
     
  4. Nov 7, 2013 #3
    Ahhhh, sorry I'm wrong... Hopefully I can be right about this:

    The equation for the bullet's X position would be X = (Vx)(Vy/9.8m/s^2), the bullet's Y position would be y = ((Vyo+Vyf)/2)(Vy/9.8m/s^2), and the equation for the monkey's vertical position is y = (Vo+Vi)/2)(Vf/9.8m/s^2)


    I hope that is correct.. Thank you for taking the time to help me :).
     
  5. Nov 8, 2013 #4
    Obviously the bullet wouldn't hit the monkey (if didn't drop) if it was pointed directly at it as gravity would pull it down, so it wouldn't be direct, but I'm not really sure how to show this with algebra..
     
  6. Nov 8, 2013 #5

    cepheid

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    Expressions for the vertical position vs. time of the bullet and the monkey should be exactly the same.

    Regarding your equations: you seem to have substituted some things in for t. I'm not sure why, to start, you just don't keep things in terms of t. For instance, the x-position of the bullet is x = vxt since this is motion at a constant speed. You substituted in t = vy/g, but I think you should wait until you have all your equations set up before deciding what needs to be substituted in. Similarly for your expressions for the y-position: for these just write down the standard kinematics equations for distance vs. time given constant acceleration.
     
  7. Nov 9, 2013 #6
    I figured out the first part!!! Displacement of bullet at Dx is Vo x Sin 0 x t - 1/2gt^2 which equals Dy - 1/2gt^2 and the displacement of the monkey is Dy - 1/2gt^2... which is the same as the bullet so they meet! I don't understand how you determine the minimum speed though, but I get the bullet's velocity's vertical must be great enough not to hit the ground until it reaches Dx and the horizontal component must be fast enough to reach the monkey before it hits the ground
     
    Last edited: Nov 9, 2013
  8. Nov 9, 2013 #7

    haruspex

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    Where did you get this equation from? Vo = (Dx/cosθ)(sqrt(g/2Dy))
    It gives the answer. If you want to get it without using that equation, consider what happens if the bullet is fired at the minimum speed. Where will the monkey be when the bullet hits it? What equations can you write involving t, Dx, g, θ and v to express that?
     
  9. Nov 9, 2013 #8
    I got the equation Vo = (Dx/cosθ)(sqrt(g/2Dy)) from my teacher, but he says I need to derive it. At the minimum speed, the monkey and the bullet will meet at (Dx,0).. I'm not sure how to fit the angles into the equation though... How do Sinθ and Cosθ play into it? And thank you guys so much for taking the time to help me
     
  10. Nov 9, 2013 #9

    haruspex

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    OK. In that equation there are three variables where two will do. That's because there's a relationship between them: tan(θ) = Dy/Dx. This means that there are several different-looking ways of writing what is really the same equation. It's not obvious to me how your teacher picked that particular form, so let's try to derive at least one version of it.
    Consider the trajectory of the bullet. If it travels for time t, what equations can you write for where it will be (x and y)? What do you find if you plug in that final position and eliminate t?
     
  11. Nov 10, 2013 #10
    Time should equal t = 2((Vo)(Sinθ))/g

    Position for Dx should be (VoCosθ)(2((Vo)(Sinθ))/g) And Position for Dy should be VoSinθ(2((Vo)(Sinθ))/g, but won't the Dy position be 0?

    How would I make this an equation? I feel so stupid...
     
  12. Nov 10, 2013 #11

    haruspex

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    Right. So rearrange that in the form Vo = ...
    Dy is the original height of the monkey, not its height when it lands. Anyway, you have enough with
    Dx = (VoCosθ)(2((Vo)(Sinθ))/g) to get one form of the equation you're after, and you can use that to answer (b). Do you need to derive exactly the form in the OP?
     
  13. Nov 10, 2013 #12
    Okay so I derived Vo = sqrt((0.5(Dx/cos0/sinO))g)

    Correct? And yeah I need to derive it exactly like he did... Somehow -.- .

    Anyways.. I feel like I'm getting somewhere! Thank you for helping me
     
  14. Nov 10, 2013 #13

    haruspex

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    An easy way to do that is simply equate your expression with his and simplify until you get down to some known fact. Then you can reverse the steps to show the two expressions are equivalent.
     
  15. Nov 10, 2013 #14
    Hmm, do you know any other ways I could derive Vo?
     
  16. Nov 11, 2013 #15

    haruspex

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    Let's try picking the target equation apart:
    Vo = (Dx/cosθ)(sqrt(g/2Dy))
    Vo cosθ = Dx(sqrt(g/2Dy))
    The LHS is the horizontal velocity. Express that in terms of Dx and t. Can you also write t in terms of g and Dy?
     
  17. Nov 11, 2013 #16
    I don't know where he is deriving sqrt(g/2dy$
     
  18. Nov 11, 2013 #17

    haruspex

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    If you followed what I wrote last time, you'll know we need to show 1/t = sqrt(g/2dy).
    So let's unwind that:
    t2 = 2Dy/g
    gt2/2 = Dy
    This is now looking very familiar. Can you see why it would be true?
     
  19. Nov 12, 2013 #18
    Wow! I totally didn't see you could sub Dy in for VySin0... So Dx = VoCos0(sqrt(g/2dy),
    So Vo would equal Vo = (Dx/Cos)/(sqrt(g/2dy)). I just know I am wrong...
     
  20. Nov 12, 2013 #19

    haruspex

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    Vy Sinθ t
    Something inverted there. sqrt(g/2dy) has units 1/time.
     
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