Wondering where the distance formula from acceleration due to gravity comes from

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The distance formula d=1/2gt^2 derives from integrating the acceleration due to gravity, represented as a constant g. To obtain this formula, one must integrate the acceleration function twice with respect to time, which introduces two arbitrary constants of integration. Solving the initial value problem with conditions y(0)=0 and y'(0)=0 leads to the specific solution for distance fallen under gravity. This process highlights the relationship between acceleration, velocity, and displacement in the context of free fall. Understanding this derivation is essential for grasping the fundamentals of motion under constant acceleration.
Icedfire01
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Pretty much like the title says. I'm having a hard time finding where the formula: d=1/2gt^2 comes from. Any help would be greatly appreciated.
 
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If a=g, then integrate twice with respect to time !
 
That's almost right. When you integrate twice you pick up two arbitrary constants of integration. To really derive the formula you would have to solve the following initial value problem:

\frac{d^2y}{dt^2}=-g

y(0)=0

y'(0)=0.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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