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

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SUMMARY

Galileo's Law of Free Fall is represented by the formula y = 16t², where y is the distance traveled after time t in seconds. The constant 16 is derived from the acceleration due to gravity, which is 32 feet/sec², and is halved in the formula to account for the average speed during free fall. This law applies specifically to objects dropped from a height not exceeding certain limits, as acceleration can vary with height. The factor of 0.5 in the formula arises from solving the differential equation that describes the motion of a free-falling body.

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  • Understanding of basic calculus concepts, particularly differentiation.
  • Familiarity with the physics of motion, specifically free fall and acceleration.
  • Knowledge of the relationship between distance, speed, and time in kinematics.
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BogMonkey
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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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