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An object whose mass is $m$ is crashing into the ground and leaping with the same velocity. If the duration of the interaction between the object and the ground is $0.1$ seconds, determine the force ground applied to the object.

I know that

$$\Delta P_y = F_y \times \Delta t$$

$$\Delta P_y = (N-mg) \times \Delta t$$

$$2mv_0 \sin \theta = (N-mg) \times \Delta t$$

Hence we get

$$N = 220N$$

According to answer key, it seems incorrect.

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  • $\begingroup$ Is there anyone who can take a look at it? $\endgroup$ – Busi May 16 '18 at 19:44
  • $\begingroup$ I'm still having trouble... $\endgroup$ – Busi May 16 '18 at 20:09
  • $\begingroup$ mg plays no role here. $\endgroup$ – user58697 May 16 '18 at 20:20
  • $\begingroup$ @user58697 I realized that. $\endgroup$ – Busi May 16 '18 at 20:21
  • $\begingroup$ Really, Isn't there anyone? $\endgroup$ – Busi May 16 '18 at 20:41
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Usually this is an impulse force which is triangular in shape. However, assuming its a constant force (rectangular), then it should be:

The normal velocity is $5m/s$ to the ground. It reduces to $0 m/s$ in $.05$ seconds. So $F = ma$ therefore: $$ F = 2\cdot \frac{5}{.05} = 200 N$$

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