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

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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