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A body with mass $m$ is released from the height $h$ without initial velocity. Also i assume the impact is wholly plastic upon impact with the ground ( at $ x=0 $ ) i want to calculate the forces acting on it.

I will first present my way :

1.Integrating acceleration
$a(t)=-g$
$v(t)=-g*t$
$x(t)=-\frac 12*g*t^2+h$

2.Get the velocity shortly/during before impact $x=0$ ,so
$ t_{impact}= \sqrt{\frac{2*h}{g}}$ and $v_{impact}=-g*\sqrt{\frac{2*h}{g}}$

3.Now i use conservation of momentum. First because of my plasticity assumption, the velocity after the impact is 0.

$-m*v_{beforeImpact} = \int_{0}^{t_{impact}} F = \bar{F} $

i assume $v_{impact} = v_{beforeImpact} $ and get $m*g*\sqrt{\frac{2*h}{g}} = \bar{F} $

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Now my Questions :
1. First is this correct ?
2. How would i average $\bar{F} $ ? Deviding it by $t_{impact}$ ?But it would go to infinity if the $t_{impact}$ goes to zero. 3. How would i calculate the peak force ?

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    $\begingroup$ You can tell pretty easily that -m*v = ... = F isn't right. kgm/s isn't the same unit as kgm/s^2. $\endgroup$ – Ulthran Aug 11 '16 at 13:39
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Your work is correct up until this line: $$mv = \int_0^{t_\textrm{impact}}F\,\mathrm dt = \bar{F}$$ There are two mistakes here. First, the end result of the integral in terms of the average force is $$\int_0^tF\,\mathrm dt = \bar{F}\Delta t$$ Second, the $t_\textrm{impact}$ in the integral limit is different from the $t_\textrm{impact}$ in your velocity equations. In the equation $t_\textrm{impact} = \sqrt{\frac{2h}{g}}$, the time refers to how long the object falls before it hits the ground. The $t_\textrm{impact}$ in the integral is how long the collision with the ground lasts. These should be given different names to avoid confusion like $t_\textrm{fall}$ and $t_\textrm{collision}$.

As to how to calculate the force of impact, that depends on the specific object being dropped. You can imagine that a water balloon going splat against the ground would have a large $t_\textrm{collision}$ because it spends a lot of time deforming and flattening out on impact (see this video). This means that the force of impact is small. On the other hand, a baseball would have a very short $t_\textrm{collision}$ because it doesn't deform on impact and stops almost immediately, resulting in a large force.

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  • $\begingroup$ The conclusion from this: you can't calculate the force unless you know the duration of the collision. You (the OP) don't have enough information. $\endgroup$ – garyp Oct 1 '16 at 14:40

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