GREPhysics.NET: GR0177 #24

GR0177 #24

Problem

GREPhysics.NET Official Solution

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Mechanics $\Rightarrow$ }Kinematics

If a stone is thrown at such an angle at an initial velocity, its horizontal $v_x$ vs t graph should be constant and positive $v_x=v_{x0}=v_0\cos(45^\circ)$ . Thus, choices (A) and (E) are out.

Recalling the basic kinematics equation $v_y=v_{y0}-gt$ , one eliminates choice (D), since that shows a parabolic time dependence, when a linear one is required. Since the slope is negative, the $v_y$ -graph should look like III one has choice (C).

(If one forgets the basic equations above, one can derive it all from summing up the net force $\ddot{y} = -g$ . Integrate both sides to get velocity. Integrate again to get position.)

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Comments

johnhero2010 2012-10-08 02:45:31	Since air resistance is being ignored, there is no force in the horizontal direction. Therefore, the x-component 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 y-component 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. Reply to this comment
secretempire1 2012-08-28 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. Reply to this comment
isentropic 2008-10-17 08:24:26	You could also plot the parabolic trajectory of the stone on an xy-plane and draw the velocity components. From there, it should be pretty easy to determine what the vt-graphs should look like. Reply to this comment
antithesis 2007-10-01 11:54:01	Actually, in this case, if you only look at $v_y$ , and realize it is case III, the only answer is (C), and you don't even need to consider $v_x$ Reply to this comment

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You could also plot the parabolic trajectory of the stone on an xy-plane and draw the velocity components. From there, it should be pretty easy to determine what the vt-graphs should look like.

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