GREPhysics.NET: GR0177 #24

GR0177 #24

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

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.)

Alternate Solutions

There are no Alternate Solutions for this problem. Be the first to post one!

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

Post A Comment!

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone