Deviation of Newton's Gravity Equations

[eNathan]

$$T = \sqrt { \frac{2d} {g} }$$
Therefore (and I had to come up with this on my own because I could not find it on the internet :rofl:)
$$D = \frac{t^2 * g} {2}$$

But I cannot find the logic in why the time equals 2 times the distance over the acceleration of gravity on earth. Now, I do understand the units. The Time is in seconds, and the Distance is in meters, because the acceleration is in meters per second (any other units will work just fine).

But why is the distance multiplied by 2?
And why is the square root function used?

 

[dextercioby]

Do you know calculus...?

Daniel.

 

[eNathan]

hehe, I only wish I did. I am researching it in my spare time, and I plan to take classes on it but in shorts no I don't know calc.

But it won't hurt if you try to explained it to me anyway if that was your plan :)

 

[dextercioby]

To integrate Newton's second law in scalar form

$$\frac{d^{2}y(t)}{dt^{2}}=g$$

subject to the initial conditions

$$y\left(t_{0}\right)=y_{0}$$

$$\frac{dy}{dt}\left(t_{0})=:v\left(t_{0}\right)=v_{0}$$


Daniel.

 

[whozum]

More simply, I'm sure you know the position equation for an object in constant acceleration:

$$x = x_0 + v_0t + \frac{gt^2}{2}$$

Notice if [itex] x_0 = v_0 = 0 [/tex] then solving for t:

$$t = \sqrt{\frac{2x}{g}}$$

Your equation is only valid when the initial velocity is zero.

 

[eNathan]

More simply, I'm sure you know the position equation for an object in constant acceleration:

$$x = x_0 + v_0t + \frac{gt^2}{2}$$

Notice if [itex] x_0 = v_0 = 0 [/tex] then solving for t:

$$t = \sqrt{\frac{2x}{g}}$$

Your equation is only valid when the initial velocity is zero.

$$x = x_0 + v_0t + \frac{gt^2}{2}$$

Hmn, I think I get your point, but why do you square numbers all the time anyway? I mean, I see it everywhere. e=mc^2 ^2 ^2 blah every equation in physics involves $$x^2$$ :rofl: Maybe if you explain why everything need be squared, I can understand it

Thx

 

[arivero]

$$T = \sqrt { \frac{2d} {g} }$$
Therefore (and I had to come up with this on my own because I could not find it on the internet :rofl:)
$$D = \frac{t^2 * g} {2}$$

But I cannot find the logic in why the time equals 2 times the distance over the acceleration of gravity on earth. Now, I do understand the units. The Time is in seconds, and the Distance is in meters, because the acceleration is in meters per second (any other units will work just fine).

Hope it is a type, and you use meters per second squared for the acceleration. Also, this thread should be titled "Deduction (or derivation) of Galilean law for free falling bodies"; Newton has no role here except to confirm us that acceleration g is approximately constant at Earth surface.

Indeed the original proof of the time square formula can be found in internet here:
http://www.mpiwg-berlin.mpg.de/Galileo_Prototype/DHTML/D202.HTM

 

[arivero]

Indeed the original proof of the time square formula can be found in internet :: here:
http://www.mpiwg-berlin.mpg.de/Galileo_Prototype/DHTML/D202.HTM

This requires a bit of history. In order to get students, Galileo did not publish the parabolic trajectory until Cavalieri did (rightly attributing it to master Galileo). Then he become furious for a moment but finally he thought better and he finished his http://oll.libertyfund.org/ToC/0416.php in a notebook of Galileo, mss 72, probably used during the composition of the book.

 

[whozum]

$$x = x_0 + v_0t + \frac{gt^2}{2}$$

Hmn, I think I get your point, but why do you square numbers all the time anyway? I mean, I see it everywhere. e=mc^2 ^2 ^2 blah every equation in physics involves $$x^2$$ :rofl: Maybe if you explain why everything need be squared, I can understand it

Thx

The reason is calculus, and definitions of acceleration, velocity, and position.

Acceleration is the rate of change of velocity:

[itex] a = \frac{dv}{dt} [/tex] so then [itex] \int{a}{dt} = \int{dv} [/tex]

Integral of a constant acceleration "a" is just 'a' times the dependant variable, t. v = at+v_0. v_0 is the integration constant.

$v = at+v_0$

Velocity is the rate of change of position:

$v = at+v_0 = \frac{dx}{dt}$ so then [itex] x = \int{(at+v_0)}{dt} [/tex]

Integral of a linear function of time 'at' is 'at^2/2':

$$x = \int{at+v_0}{dt} = v_0t+\frac{at^2}{2} + x_0$$