Jump to content

Thrown at different angles, from same height...


MDJH
 Share

Recommended Posts

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?

Link to comment
Share on other sites

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...

Link to comment
Share on other sites

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 by darkenlighten
Link to comment
Share on other sites

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.

Link to comment
Share on other sites

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 merged
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.

 

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?

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

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?

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.