GR0177 #24



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Comments 
johnhero2010 20121008 02:45:31  Since air resistance is being ignored, there is no force in the horizontal direction. Therefore, the xcomponent of the particle's velocity is constant. We are told that the stone is thrown in the +x direction. Therefore, the ( v_x) component of the velocity must be positive. Therefore, graph II represents
(v_x) versus (t) . The ycomponent of the particle's velocity vector is initially positive, but it decreases at a constant rate due to the force of gravity, and eventually becomes negative. Therefore, graph III represents (v_y) versus (t) . Therefore, answer (C) is correct.
  secretempire1 20120828 08:13:12  If you consider just the vertical velocity of the stone, this problem becomes very simple.
You initially toss the stone up, so the y velocity is initially some positive number. This rules out I and V. As gravity acts on the stone, it gets slower and slower and slower until it reaches zero. Then it starts falling, giving it a negative y velocity.
The only graph that represents this behavior is III. And only choice C offers graph III as the choice for the y velocity.   isentropic 20081017 08:24:26  You could also plot the parabolic trajectory of the stone on an xyplane and draw the velocity components. From there, it should be pretty easy to determine what the vtgraphs should look like.   antithesis 20071001 11:54:01  Actually, in this case, if you only look at , and realize it is case III, the only answer is (C), and you don't even need to consider  

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