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Monkey and hunter - forces, vectors

  1. Dec 28, 2013 #1
    Monkey and hunter -- forces, vectors

    1. The problem statement, all variables and given/known data

    I'm currently supplementing my institutional education with some me-time study. I'm studying maths and physics A-levels (or, rather, A-level equivalent).

    Yale University has an awesome "online courseware" which makes available to the public videos of lectures and study materials (problem sets, etc.). My problem refers to the problem set given for the Newton's Laws lectures, found here: problem set, solutions, specifically problem 8. The problem is quoted here:

    2. Relevant equations

    I began with asking what are the positions for each object's motion. The monkey is easy. It's "x" position is unchanging, and so does not vary with time: [tex]x_m =d [/tex]; while it's "y" position is changing with time: [tex]y_m = h-\frac{1}{2} g t^2[/tex]. Next, and this is where my approach differs to that given in the solution, I define the components of the bullet. The bullet begins with some initial velocity [tex]v_0[/tex] in the horizontal direction, and so by my understanding of forces and newtonian mechanics, this horizontal velocity is constant (ignoring air resistance), so that the bullet's "x" position is: [tex]x_b=v_0 t[/tex]; and like the monkey, the bullet's "y" position changes because of gravity: [tex]y_b=h-\frac{1}{2} g t^2[/tex].

    3. The attempt at a solution

    My reasoning is then: for the bullet to hit the monkey, the bullet must travel a horizontal distance d in a time that is no greater than the time it takes the monkey to fall a distance h; and because I'm dealing with the horizontal distance, I can focus on the "horizontal" component of the bullet's trajectory. Thus, the time for the bullet to travel a distance d is: [tex]t_{b}=\frac{d}{v_0}[/tex]; and the time it takes for the monkey to reach the ground is (given by solving [tex]y_m=0[/tex] for t): [tex]t_m=\sqrt{\frac{2h}{g}}[/tex]. Then equating the two and solving for [tex]v_0[/tex]: [tex]v_0=\frac{d\sqrt{g}}{\sqrt{2h}}[/tex].

    Can someone spot the error in my reasoning? Many thanks!

    P.S. Sorry for the poor formatting. I'm not familiar with this forum's LaTeX tools.
  2. jcsd
  3. Dec 28, 2013 #2


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    They seem to have forgotten to mention that on hearing the shot the monkey lets go of the branch (and the bullet is much slower than the speed of sound). (And you lost the "v >" in quoting the condition.)
    ... which is the correct answer for what you focused on - the horizontal component. But the question posed wants the total speed.
  4. Oct 30, 2017 #3
    I have a similar question: If somebody wants to shoot an object falling from a cliff with a parachute attached, where should he aim?
    There is a drag force involved and that's why I think it is better to aim above the package. Drag force increases proportionally to velocity squared.
    Am I right?
  5. Oct 30, 2017 #4


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    To figure out where to aim you would need to solve the drag equation. If it is directly proportional to the square of the speed then you can certainly find the velocity as a function of time (need to integrate a secant), but as I recall, the second integral to find the displacement by time can only be done numerically.
    In reality, it's more like a general quadratic, involving a linear term.

    The bullet also suffers drag. This is way more complicated because it is not moving vertically.
  6. Oct 30, 2017 #5
    What the question may be asking is to show that
    h - 1/2 g t^2 > 0
    Doesn't this have to be true if the bullet is to reach the monkey before the monkey reaches the ground?
    Now use this to eliminate t from the equation using the given conditions.
  7. Oct 30, 2017 #6


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    The original question was dealt with 4 years ago. The resurrection of the thread concerns drag. Probably should have been a new thread.
  8. Oct 30, 2017 #7
    In the monkey and hunter problem, the traditional explanation is that the bullet falls away
    from the initial line of sight at the same rate that the monkey falls.
    Thus, the bullet will hit the monkey.
    Now, qualitatively, it should be clear that if the monkey falls at a slower rate than the
    bullet then the bullet would pass under the monkey if the hunter was initially
    aiming directly at the monkey.
  9. Oct 30, 2017 #8


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    Yes, that's what Poetria deduced in post #3.
  10. Oct 31, 2017 #9
    This question is a kind of thought experiment - too little data. At first, I thought it would be better to aim below the object with a parachute - I didn't take into account gravity acting on the arrow and I thought drag force would be zero at the instant of shooting as initial velocity is zero.

    Sorry for having caused 'drag' on the forum. :) I was searching for similar situations to understand this better. :)
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