But I proved that \(-5t^2\neq x\) on Earth, didn't I?
Also, \(g\) is used for gravitational acceleration.
A force is in newtons, which is equivalent to \([kg\ m/s^2]\). But, for that equation, you want gravitational acceleration, the acceleration. This is the acceleration due to gravity, which is a force. Now, that is all semantics, but it doesn't not alter my proof! Is my proof correct?
As I said, you can use different approximations for gravitational acceleration. The \(10\ [m/s^2]\) is one common approximation.
I could not find your second equation at the link I posted. However, there is the first one that is \(d=\frac{1}{2}g\ t^2\), which implies that \(t^2=\dfrac{2\ d}{g}\) and so \(-5t^2=\dfrac{-10g}{d}\). Now, if it is true that \(-5t^2=\dfrac{-10g}{d}\), and that \(-5t^2=d\), then \(\dfrac{-10g}{d}=d\). This implies that \(-10g=d^2\). Since \(-10g\) is nearly constant of a falling object that we are describing, then \(d^2\) must also be constant, even with changing time and a non-zero acceleration. That is not true - the square of the distance i very obviously changing. Assuming Wikipedia's equation to be correct, and the reasoning that the square of the distance will not be equal to the opposite of the acceleration increased by a factor of 10, then your equation is incorrect.
Which assumption of mine would be wrong? I just don't want a flawed equation to be on OpenStudy, so please show me the error in my earlier proof. This last one is not an actual proof, because it is true only assuming that \(-10g=d^2\). This is because gravitational acceleration changes with distance, though, and it takes a considerable distance to make a slight change in gravitational acceleration.