20101028, 23:55  #1 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
true or false
could it ever be true that going down a steeper slope will get you going slower than a shallow slope for the same vertical distance. if you agree it's true can someone send it to mythbusters to test ? I don't know quite how to. though I might try to find adam savages email.

20101029, 11:30  #2 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17·251 Posts 
I don't think it's possible. If you want to suggest it, you can by posting on this forum.

20101029, 11:45  #3  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
37×281 Posts 
Quote:
Paul 

20101029, 11:57  #4  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20110504, 13:49  #5  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:


20110504, 17:34  #6 
"William"
May 2003
New Haven
2^{3}·5·59 Posts 
Assuming you don't want to rig the results using methods such as those suggested by Xilman above, you are converting gravitational energy into kinetic energy. If there aren't any energy leaks (friction etc), then the speed must be the same.

20110504, 18:42  #7  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
9.808 angle 89 using sine this is what I come up for for 9.80665 m drop on the opposite ( admittedly doesn't give 9.80665 m in 1 second I know). Last fiddled with by science_man_88 on 20110504 at 18:43 

20110504, 19:01  #8 
May 2003
11000001011_{2} Posts 
It also depends on what you mean by slope.
For example, your track could consist of two metal wires. The moving object could be two cones glued together. Then it can be the case that you can have your cones move "up" the track. 
20110510, 01:48  #9 
Dec 2010
Monticello
5×359 Posts 
Guys:
There's also an assumption of linearity here...for example, if the thing going down the slope is a rock, and at the end of the slope there's a water surface, then, if the angle is too steep, the rock falls into the water. If it is shallower, then the rock will skip, and bounce back into the air with little loss of speed. Similarly, at work we have a selfexciting brake circuit, that, basically, doesn't "catch" until a certain speed is reached. Get above that speed, you end up much slower than if you stayed just below it. 
20110510, 12:37  #10  
Nov 2003
2^{2}×5×373 Posts 
Quote:
If the surface is frictionless and the object falls through height 'h', then its total velocity will be 'v' as given by mgh = 1/2 mv^2 where m is the mass and g the force of gravity. But part of that velocity will be vertical and part of it will be horizontal. Total velocity will be independent of slope, but VERTICAL velocity will not be. The question was about vertical velocity. Thank you for playing. If the slope has angle theta (with the ground), I suggest you resolve the velocity into its horizontal and vertical components. (I assume you know what a dot product is??? [actually, dot products can be avoided here]. You will find that as theta > 0 (the slope becomes flat), the VERTICAL velocity also goes to 0...... (Think about sin(x)). I assume of course, a frictionless slope. If it is not, then one must also compute the frictional force (ratio of vertical and horizontal force against the surface of contact). This will have the effect of opposing 'g', the force of gravity. And the frictional force will depend on the slope. This is all simple high school physics. 

20110510, 12:58  #11  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
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