Does anybody have a paper on Galileo's law of falling bodies y=16t'2 ?

stabatpatriae said:
Why do we take 16 as constant and why do we square t?
Click to expand...
Assuming that this question is arising from an assignment in a "class":

What is the name of the course (general science, or physics or math, etc.) and what level (highschool, first year college)?

Next question - Have you studied:

Motion of a particle under constant acceleration?
 
Have you studied any Calculus yet? This will be extremely helpful in answering your question.
 
stabatpatriae said:
Why do we take 16 as constant and why do we square t?
Click to expand...
It's not magic. It's based off empirical study and mathematical modeling.

The model (or formula) is built for Constant Acceleration. That's why 16 is constant.
It's 16 (or -16) because that's what fits the model in ft/sec.
t is squared because that's what fits the model.
 
Subhotosh Khan said:
Assuming that this question is arising from an assignment in a "class":

What is the name of the course (general science, or physics or math, etc.) and what level (highschool, first year college)?

Next question - Have you studied:

Motion of a particle under constant acceleration?
Click to expand...
I was looking to my screenshots in my phone and there was this formula I probably took from a book I unfortunately can't remember the name of. I haven't studied the topix you mentioned.
 
stabatpatriae said:
I know basics of functions and limits and derivatives
Click to expand...
OK, great! ay=acceleration = 32ft/sec2
Now the derivative of vy=velocity wrt to time, t, is acceleration. So what is vy?
The derivative of the position vector, y, is the velocity vector. So what is y?
 
IF you know basic Calculus, then

An object's velocity is the derivative of its position function: v= ds/dt.
An object's acceleration is the derivative of its velocity function: a= dv/dt.

Gallileo measured a falling objects acceleration as the constant -32 feet per second per second. (Well, he didn't use "feet" or "seconds" but got the equivalent in whatever units he used.)

dv/dt= -32 so, dv= -32dt and, integrating, v= -32t+ C feet per second where C is the beginning speed. If the object is dropped from rest, C= 0 so v= -34t.

So v= dx/dt= -32t and then dx= -32t dt. Integrating again, \(\displaystyle x= -16t^2+ C\) where this C is the height from which the object is dropped. If we take that to be 0 then \(\displaystyle x= -16t^2\).
 
If I remember correctly, that is discussed in Galileo's book, Two New Sciences
 
HallsofIvy said:
IF you know basic Calculus, then

An object's velocity is the derivative of its position function: v= ds/dt.
An object's acceleration is the derivative of its velocity function: a= dv/dt.

Gallileo measured a falling objects acceleration as the constant -32 feet per second per second. (Well, he didn't use "feet" or "seconds" but got the equivalent in whatever units he used.)

dv/dt= -32 so, dv= -32dt and, integrating, v= -32t+ C feet per second where C is the beginning speed. If the object is dropped from rest, C= 0 so v= -34t.

So v= dx/dt= -32t and then dx= -32t dt. Integrating again, \(\displaystyle x= -16t^2+ C\) where this C is the height from which the object is dropped. If we take that to be 0 then \(\displaystyle x= -16t^2\).
Click to expand...
Thank you so much!
 
You must log in or register to reply here.
Top