MDJH Posted June 19, 2010 Posted June 19, 2010 Treating air resistance as negligible for this case, suppose we have three objects thrown FROM one vertical position TO another vertical position; the initial and final vertical positions being the same for each object. They are thrown at the same speeds, but at different velocities; one is thrown at an upward angle, one is thrown at a downward angle, and one is thrown horizontally. Would they have the same speeds, or different speeds, upon hitting the ground?
swansont Posted June 19, 2010 Posted June 19, 2010 Try applying conservation of energy and answering this.
MDJH Posted June 20, 2010 Author Posted June 20, 2010 Try applying conservation of energy and answering this. Well, should I assume that gravitational potential energy and macroscopic kinetic energy are the only forms of energy involved in this case? If so they'd have to land at the same speed, but I'm wondering if there's something I'm missing...
cypress Posted June 20, 2010 Posted June 20, 2010 No, you are not missing anything... Sounds good to me.
darkenlighten Posted June 20, 2010 Posted June 20, 2010 (edited) Unless I'm thinking about this wrong, they will not have the same final velocities, they will be different. I think the thing here is the fact that they will indeed fall as the same rate ( acceleration = g) , but the direction of their velocity is going to effect their final velocities. For example, Labeling (1) Upward at an angle, (2) downward and (3) horizontally...you want to consider only their vertical velocities, which they are not all the same, (1) and (2) only have the same velocities when heading down if (1) is thrown straight upward and (3) always has a 0 downward velocity. So according to this they will not all have the sames velocities when hitting the ground since [math] v_f = v_0 + at [/math]. Unless I'm crazy. Edited June 20, 2010 by darkenlighten
timo Posted June 20, 2010 Posted June 20, 2010 The question is not asking for equal velocities upon impact but for equal speeds, i.e. magnitudes of velocities. You are of course be correct about different velocities: the horizontal part of the velocity is differs when thrown and not change during the flight. Note that the question is very homework-like, so I think you should be a bit reluctant to discuss details of a solution until MDJH has presented a solution or at least a week or so has passed.
swansont Posted June 20, 2010 Posted June 20, 2010 Note that the question is very homework-like, so I think you should be a bit reluctant to discuss details of a solution until MDJH has presented a solution or at least a week or so has passed. Agreed, but I think that MDJH has done so in post 3. Merged post follows: Consecutive posts mergedUnless I'm thinking about this wrong, they will not have the same final velocities, they will be different. I think the thing here is the fact that they will indeed fall as the same rate ( acceleration = g) , but the direction of their velocity is going to effect their final velocities. For example, Labeling (1) Upward at an angle, (2) downward and (3) horizontally...you want to consider only their vertical velocities, which they are not all the same, (1) and (2) only have the same velocities when heading down if (1) is thrown straight upward and (3) always has a 0 downward velocity. So according to this they will not all have the sames velocities when hitting the ground since [math] v_f = v_0 + at [/math]. Unless I'm crazy. You can investigate the 2 vertical cases without worrying about horizontal components. If you send an object vertically at v_0, what is its speed when it hits the ground?
MDJH Posted June 20, 2010 Author Posted June 20, 2010 The question is not asking for equal velocities upon impact but for equal speeds, i.e. magnitudes of velocities. You are of course be correct about different velocities: the horizontal part of the velocity is differs when thrown and not change during the flight.Note that the question is very homework-like, so I think you should be a bit reluctant to discuss details of a solution until MDJH has presented a solution or at least a week or so has passed. I'm not asking for help with homework, (well, not directly anyway) though I can understand why you got that impression. I did introductory physics courses during my first year of university, (didn't learn that thoroughly though) as well as a few more physics courses since, (again, didn't learn it very thoroughly) but I've been unsure about what to major in for a while and have been meandering with all kinds of introductory courses for subjects other than physics; but I've been considering returning to a physics major, and was just trying to refresh my memory on this kind of stuff, thus I'm going through my first year physics book looking for whatever things I don't get, or am not sure if I get, so as to check my understanding.
darkenlighten Posted June 20, 2010 Posted June 20, 2010 (edited) Their speeds will still be different when hitting the ground, due to only the vertical distances and starting velocities (think of potential and kinetic energies if you like) Edited June 20, 2010 by darkenlighten
MDJH Posted June 20, 2010 Author Posted June 20, 2010 Their speeds will still be different when hitting the ground, due to only the vertical distances and starting velocities (think of potential and kinetic energies if you like) What do you mean?
swansont Posted June 20, 2010 Posted June 20, 2010 I think that's better phrased as: you can only get different final speeds if you have different initial speeds, or different heights.
... Posted June 20, 2010 Posted June 20, 2010 I agree with the assessment that the speeds will be the same and velocities different. If released at the same vertical position (same potential E) with the same speed (same kinetic E) and arrive at the same final vertical position (same final potential E) ... for energy to be conserved the final kinetic Es will need to be the same as well which means the velocity magnitudes (speeds) will be the same...assuming of course they all have the same mass.
J.C.MacSwell Posted June 21, 2010 Posted June 21, 2010 If it's high enough to reach terminal velocity... they will all approach it. Edit: missed the "negligible" part.
darkenlighten Posted June 21, 2010 Posted June 21, 2010 I realized I was only considering the vertical velocity components and not the velocity vector...so with that from energy conservation its easy to see [math] E_o = E_f \Rightarrow -mgh_o + \frac{1}{2}mv{_o}{^2} = -mgh_f + \frac{1}{2}mv{_f}{^2} \Rightarrow v_f = [v{_o}{^2} + 2gh_f]{^ \frac{1}{2}}[/math] So since [math] v_o [/math] and [math] h_f [/math] are the same for all so yea that's it. Same speed! And to add to "...", mass of the objects will not affect anything. They could all have different masses and nothing will change.
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