What is the Constant of Proportionality in Galileo's Law of Free Fall?

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Galileo's Law of Free Fall is expressed as y = 16t^2, where y represents the distance fallen after time t, and 16 is the constant of proportionality derived from the acceleration due to gravity. This constant applies specifically to scenarios where the time is measured in seconds and the distance in feet. The general equation for free fall is y = 0.5at^2, with 'a' being the acceleration, which is 32 feet/sec^2 on Earth. The factor of 0.5 arises from solving the differential equation for a free-falling object. Understanding these constants is crucial for accurately calculating the motion of falling objects.
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I'm reading a maths book called Thomas Calculus and in their method for getting the average speed of an object when only the height its dropped from is known is this formula here which they call Galileos law:
y = 16t^2
y being the distance traveled after time. What I don't get is where they get the 16 from. All they say about it is "where 16 is the constant of proportionality". Where did they get this constant of proportionality from and does this 16 apply to all falling object scenarios?
 
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The general form would be y = .5at^2, where a is the acceleration. The acceleration due to gravity (free fall) is 32 feet/sec^2. So the formula you have been given is for time in seconds and distance in feet, when dropping an object from a height (not too large, since "a" will depend on height) above the surface of the earth.
 
Thanks that explains where they got 16 but why .5?
 
The .5 comes from solving the Differential equation which describes a free falling body.
 
And if you're reading a calculus problem, the time derivative of that position would give you y' = a*t = v which is obviously your speed at any given time, t, given a constant acceleration, a, if begun at rest.
 
Ah right. Thanks.
 
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