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When a ball is thrown at a person at certain distance "d" first with lower velocity and then with higher velocity, it is obvious that it would hurt more in second case. So, I guess we can say the force is greater in second case. But why?

I came up with an thought that change is momentum is higher in second case as velocity changes from high to zero( supposed for easier understanding). And force is change in momentum per unit time.

Well that led me to come up with another confusion: is the time of contact (required for change in momentum) same or different in both the cases?

I also thought about change is kinetic energy being equal to work done. So the work done would be higher in the second case but again got confused whether the distance $d$ of ($W$=$F$$d$) is the distance between the thrower and the person who experiences the impact or not. If it is indeed the distance between them (which tends to be constant) then I guess the force will be greater in second case according to the formula.

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  • $\begingroup$ D is the distance along which the ball decelerates.from beginning to end of impact. $\endgroup$
    – Alchimista
    Dec 21, 2017 at 9:27

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change is momentum is higher in second case as velocity changes from high to zero( supposed for easier understanding). And force is change in momentum per unit time.

Yes.

is the time of contact (required for change in momentum) same or different in both the cases?

My guess is that the time is more or less the same. At higher impact speed, the object will "sink deeper into" the soft skin and tissue but that longer distance will be passed faster. It might take more or less the same time.

This of course depends on the part of the body you throw at. Throwing at someone's head is a place on the body with thinner skin than, say, the thigh. There the bones and skull underneath the skin will soon be reached, which will abruptly stop the motion over a very short time. This is a momentum change over a very short time, corresponding to a much larger force.

In general, harder surfaces (asphalt, tiles, glass, bone...) cause much higher impact forces over shorter time than softer or more flexible surfaces (pillow, trampoline, grass field, fat tissue...).

If the person is not being hit but rather catches the object, then he can control the time himself by using joints and moving along with the motion. When catching a basketball you can stretch your arms just before reaching it and let the arms and body follow the motion backwards during the catch. All this happens to increase the time of the impact, causing more gradual momentum transfer giving less impact force.

Same happens when you land after a jump and bend in your knees. Just try to catch a ball with stretched, non-flexing arms or land with stretched, non-bending legs and you will feel the force difference.

confused whether the distance $d$ of ($W=Fd$) is the distance between the thrower and the person who experiences the impact or not.

No, it is not. $d$ is only the distance over which work od being done. No work is done while the object flies trough the air. Work is only done when there is contact. Work is done during the impact, only. So the distance we are talking about here is the distance that the body allows the object to "sink into the skin", as mentioned above.

Is the distance short then the force must be larger in order to still absorb all the kinetic energy.

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Your line of thinking is correct regarding your first point. And yes time of contact would be necessary in calculating the force.

You must have noticed that while catching a ball if you retract your hands it hurts less, that is the force is less. This is because while retracting your hands you are essentially increasing the time of contact of the ball.

Note that the time of contact here is the time until the ball stops.

So if you have identical conditions while catching both the balls, the time of contact wouldn't differ much and the ball with higher velocity would hurt more.

About your second point, d is the distance your hand moves while applying the force to the ball. Also mind that there will be no force on the ball while travelling from the thrower to the catcher, considering no air resistance.

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