Ok, when solving for the force of impact (free falling object), there are either errors in the formula, or I am missing something.

More specifically, Finding the square root of an object's velocity falling from $5 \ km$ gives me the same result as finding the square root and object falling from $5 \ km$ but with different densities, mass, weight.

As you can see, this raises significant concerns. Especially, because I ran tests to verify. I can confirm larger masses with equal density cause significantly more damage at the same speed.

This is common sense. However, the calculus is not. Below is the equation. It is separated into two parts. Part two is needed to solve part one.

Part 1) $$Σf = ma = \frac{m(ν-v)}{Δt} = mv-\frac{mv}{Δt} = p-\frac{p}{Δt} = \frac{Δp}{Δt} = Σf = \frac{Δp}{Δt}$$

  • where momentum is p = mv,
  • where momentum is represented as a vector (kg) (m/s) = kg $\times$ m / s,
  • where acceleration is a = change in velocity/change in time = v-v/change in time.

Part 2) $$ME = ME ⇒ PE = KE ⇒ mgh = \frac{m(v)}{2}⇒ gh = \frac{(v)}{2} ⇒ v = \sqrt{2gh} $$

Does anyone know where I went wrong/what is missing? I thought about applying drag force to the equation?

  • 2
    $\begingroup$ W/o Latex rendering of math it's barely readable... $\endgroup$
    – Gert
    Nov 17, 2021 at 23:30
  • $\begingroup$ Ef = force of impact m = mass a= acceleration triangle = change v = velocity t = time p = momentum $\endgroup$ Nov 18, 2021 at 2:46
  • $\begingroup$ please see the corrected equation $\endgroup$ Nov 18, 2021 at 2:57
  • $\begingroup$ I added Latex, no it should be more readable. $\endgroup$ Nov 18, 2021 at 5:38

1 Answer 1


You need to incorporate the change in mass. That shall yield you the correct answer.

  • $\begingroup$ At what point are you suggesting I incorporate the change in mass? If you are referring to the part two initial the change in mass is accounted for. If you are referring to something else please explain. $\endgroup$ Nov 18, 2021 at 2:44

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