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Monkey and hunter - forces, vectors

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  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

    haruspex

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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.
     
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