Does higher velocity increases the impact force? [duplicate]

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In this example we have an object with 100 kg that's being dropped from a certain height (h), for simplicity lets ignore Air Resistance, by calculating the Gravitational Force (F=m.g) we would obtain the following: 98 N

Now lets say our h=100m, by using the equation for a falling object we would obtain a velocity of ~44,3 m/s at the instant moment he hits the ground.

On the other hand lets say we have h=10m, in this scenario the velocity would be 14 m/s

Each object has the same force, but different velocities. Newton's third law of motion mentions that for each action there's a reaction, meaning for each force there's an opposite force.

Now once the two objects hit the ground would the impact force be the same? Or would it change and be different due to the velocity? In the end is the Gravitational Force also the Impact Force?

marked as duplicate by JMac, Jon Custer, John Rennie gravity StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 19 '17 at 7:19

At impact, the force due to gravity is $F=mg$, where we see that a larger force corresponds to a larger mass. To rectify this with your intuition, as in a brick would hurt more if it landed on my head when dropped from a higher distance -- This is true (I don't recommend experiment), but the relevant quantity is $impulse$, which you can look up, or a more careful analysis of the acceleration of the brick when it lands on something, where we see that at contact, in a very short period of time, the velocity goes from $v \to 0$, so the force experienced is: $$F = \frac{dp}{dt} = m \frac{dv}{dt} \sim m\frac{\Delta v}{\Delta t} = \frac{v-0}{t_\epsilon}$$ Where we see that the faster the brick slows down ($t_\epsilon$) the larger the force experienced by the brick (and equally and oppositely, your head), which I think is more in the vein of what you are asking.