What is the physics formula for an object going uphill?

Assuming you know the initial velocity, the mass, the angle of the hill, and there is no friction or air resistance.

 

What is the physics formula for an object going uphill?

Assuming you know the initial velocity, the mass, the angle of the hill, and there is no friction or air resistance.

 

Hey Raider.

 

 

So.....The equation you're seeking is one that will deal with what's called Frictionless Mass on Incline.

 

Here's a cool page that shows that equation and has accompanying graphics.

 

Now the bad news....A bit of Trig is involved. But only re sin. So not too bad. LOL

 

 

http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html

Hey Raider.

 

 

So.....The equation you're seeking is one that will deal with what's called Frictionless Mass on Incline.

 

Here's a cool page that shows that equation and has accompanying graphics.

 

Now the bad news....A bit of Trig is involved. But only re sin. So not too bad. LOL

 

 

http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html

I want it going up hill. I have already calculated it going down hill.

I am perfectly capable of doing trigonometry.

Edited April 25, 20178 yr by Raider5678

Conservation of energy: velocity only depends on the height.

[math]mgh_1+\frac{mv_1^2}{2}=mgh_2+\frac{mv_2^2}{2}[/math]

Or dropping the mass:

[math]gh_1+\frac{v_1^2}{2}=gh_2+\frac{v_2^2}{2}[/math]

Conservation of energy: velocity only depends on the height.

[math]mgh_1+\frac{mv_1^2}{2}=mgh_2+\frac{mv_2^2}{2}[/math]

Or dropping the mass:

[math]gh_1+\frac{v_1^2}{2}=gh_2+\frac{v_2^2}{2}[/math]

So distance does not matter?

Nor angle?

I want it going up hill. I have already calculated it going down hill.

I am perfectly capable of doing trigonometry.

It's good to know you're capable.

 

My link did offer an ascending grade equation. Or what you do quaintly referred to as uphill.

 

You just had to, you know, scroll down a bit and not dismiss it outright because the first graphic depicted a descending grade equation. Or, to you, downhill.

It's good to know you're capable.

 

My link did offer an ascending grade equation. Or what you do quaintly referred to as uphill.

 

You just had to, you know, scroll down a bit and not dismiss it outright because the first graphic depicted a descending grade equation. Or, to you, downhill.

No, all equations were for going down the incline.

So distance does not matter?

Nor angle?

Not if there is no friction or other outside forces and the velocity is the only thing you want to know.

 

This can vastly simplify otherwise complex problems, such as roller coasters with loopings.

Edited April 26, 20178 yr by Bender

I derived the equation here:

[math] KE_{total} = Work_{gravity} + Work_{upward} [/math]

[math] \frac{1}{2}mv^2 = mgh. + mah cosec \theta[/math]

[math] \frac{v^2}{2}=h(g+a cosec \theta)[/math]

a is the acceleration of the body up the slipping surface.

Edited April 26, 20178 yr by Sriman Dutta

F = ma, and the mathematical definitions of the terms in kinematics

 

You can derive what you need from that.