Is this a proper physics derivation?

[spwath]

Did I mess anything up?

I had to Derive an expression of X in terms of H and Y.

This was for launching a projectile from a pendulum. h is the height of the pendulum, and y is the distance of the ball from launch position to the floor.

Like this:

 

I think thats all you need to know, did i mess up any physics?

 

[spwath]

I feel like I made a mistake on this, it doesnt seem to work.

[Photonic]

[Miguel Batista]

Yeah you made a mistake. When you are using the projectile equation, y=y0+v0y+0.5at2

H=y0 and it is not dependant on the time.

Launching a projectile from a pendulum doesn't make much sense to me but whatever floats your physocs teacher boat.

[spwath]

Yeah you made a mistake. When you are using the projectile equation, y=y0+v0y+0.5at2

H=y0 and it is not dependant on the time.

Launching a projectile from a pendulum doesn't make much sense to me but whatever floats your physocs teacher boat.

Wait, Im kind of confused here, what are you saying?

 

 

Also, not really a pendulum, like half a pendulum, that hits a bar, stopping at the bottom, launching a ball off the end./

[Miguel Batista]

Ah. So you want to calvulate the potential energy when the pendulum is at its highest using mgh, then, when it is at its lowest mgh is zero since all energy has been turned into kinetic energy 0.5mv2.

So from there you take the velocity.

Now using newton's motion formula,

Y=Y0+V0yT+0.5GT2

You calculate the time when Y is zero bearing in mind that V0y (initial velocity in tge Y axis) is zero. Then you use the above formula for X and use the time to solve it.

X=X0+V0xT+0.5AT2

Here the acceleration and X0 is zero so in truth the formula is just:

X=V0xT

You were overcomplicaring... Problem solved.

[spwath]

Ah. So you want to calvulate the potential energy when the pendulum is at its highest using mgh, then, when it is at its lowest mgh is zero since all energy has been turned into kinetic energy 0.5mv2.

So from there you take the velocity.

Now using newton's motion formula,

Y=Y0+V0T+0.5GT2

Youcalculate the time when Y is zero. Then you use the above formula for X abd use the time to solve it. You were overcomplicaring...

Wait, what is Y0?

We just learned Y=V0t +1/2gt2

 

The part that makes it more complicated, is that there are two y heights, from the table to the launch point, and from the ground to the table.

 

So the potential energy at the launch point would not be zero.

[Miguel Batista]

Now where you made a mistake:

Your initial assumption mgh=0.5at2 means thatthe energy total for the system is zero. From then on the trainwreck ensued. You were also using accelerations on different axis which should not be there. Gravity pnly affects the vertical axis not the horizontal one. And you were using velocities wrong too.

[spwath]

Now where you made a mistake:

Your initial assumption mgh=0.5at2 means thatthe energy total for the system is zero. From then on the trainwreck ensued. You were also using accelerations on different axis which should not be there. Gravity pnly affects the vertical axis not the horizontal one. And you were using velocities wrong too.

I said mgh=.5mv2, assuming that the potential energy is all converted into kinetic energy, so the potential and kinetic energys should be equal. Isnt that correct?

 

 

The only thing wrong is velocity, correct?

I used the velocity in the Y direction instead of velocity in the X direction, i just relized that. But time is correct, right?

 

That was wrong, hold on

[Miguel Batista]

Wait, what is Y0?

We just learned Y=V0t +1/2gt2

The part that makes it more complicated, is that there are two y heights, from the table to the launch point, and from the ground to the table.

So the potential energy at the launch point would not be zero.

Well, it is an easy enough cprrection (sorry on my phone) you know the height, you know the initial energy so you can calculate the velocity using the energy conservation theorem. The rest is the same. Y0 is the initial height once the ball is released.

[Miguel Batista]

I said mgh=.5mv2, assuming that the potential energy is all converted into kinetic energy, so the potential and kinetic energys should be equal. Isnt that correct?

The only thing wrong is velocity, correct?

I used the velocity in the Y direction instead of velocity in the X direction, i just relized that. But time is correct, right?

That was wrong, hold on

Ok. By parts:

First, although the energy will be the same, it is not the same at the same point in time. Yes you can use that formula to calculate the finql velocity IF all the eneglrgy is converted which is not the case since you still have a height in the end.

So, you need tocalculate the total potencial energy and then use that number in the equation Et=Ek+Ep (total energy is equal to kinetic energy plus potential energy).

Then you used velocities and accelerations wrong. Gravity fpes not affect the horizontal coordinate. So you don't have an aceleration there. The aceleration in the horizontal plane is zero.

In order to remember this, think about this problem. In a frictionless environment, you fire a gun and drop a bullet from your hand, at the same height. Which one will hit the ground first?

[spwath]

Ok. By parts:

First, although the energy will be the same, it is not the same at the same point in time. Yes you can use that formula to calculate the finql velocity IF all the eneglrgy is converted which is not the case since you still have a height in the end.

So, you need tocalculate the total potencial energy and then use that number in the equation Et=Ek+Ep (total energy is equal to kinetic energy plus potential energy).

Then you used velocities and accelerations wrong. Gravity fpes not affect the horizontal coordinate. So you don't have an aceleration there. The aceleration in the horizontal plane is zero.

In order to remember this, think about this problem. In a frictionless environment, you fire a gun and drop a bullet from your hand, at the same height. Which one will hit the ground first?

Well there will not be a height relitive to the launch point at launch, height relitive to the launch point is what determines the potential energy that will be converted to kinetic energy

I cant use any numbers, i just have to do a derivation.

Gravity wont effect horizontal speed, i know, but it will, as the pendulum works of gravity, and converts that into vertical speed. 

I changed my math, but I still think its wrong

 

Also, both bullets will hit the ground at the same time, so thats why my time equation should work.

 

 

EDIT:

You know what, with that new equation, the numbers seem to mostly work out

[Miguel Batista]

Well, if that is the case, the previpus solurion I gave you is right. So you calculate the initial horizpntal velocity, mgh=.5mv2

V=sqrt(2G*H)

Then, the equation is Y=Y0+V0y*T+0.5G*T2 which simplifies to Y=Y0+0.5G*T2

Which you solve to time which is T=sqrt(2(Y-Y0)/G

But, the Y will be zero since at the end of the trajectory the height is zero sp you get:

T=sqrt(2Y0/G)

Then, X=V0x*T

So X=sqrt(2G*H*2Y0/G)

Which is X=sqrt(4H*Y0) or

X=2sqrt(H*Y0)

There you go. You had it almost right.

[Tsuki]

see, your problem is that you forgot that E=mc²