$F=ma$. Falling from any distance, mass stays the same, and acceleration due to gravity stays the same. So, why does it hurt more, the longer you fall?
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asked May 9, 2015 at 9:24
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4$\begingroup$ The longer you fall, the faster you get, the more kinetic energy you possess. That energy has to go somewhere $\endgroup$– sagittarius_aMay 9, 2015 at 9:27
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2$\begingroup$ and the relevant quantity when hitting the ground is not acceleration but deceleration, from a velocity v to zero in a microsecond or so. $\endgroup$– anna vMay 9, 2015 at 9:47
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$\begingroup$ If you know about integration, one can show that, for a constant force, your kinetic energy when you hit the ground is proportional to the square of the duration of the fall. $\endgroup$– DemostheneMay 9, 2015 at 10:02
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3 Answers
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Yes, $F=ma$, but also $v=at$. That means that, as you fall for a longer time, your speed will increase.
After 1 second, you are going at $9.8 m/s$ or $35 km/h$, about the speed of Usain Bolt. After 10 seconds you would reach $98 m/sec$ or $350 km/h$. For a free-falling human, the air resistance actually limits you to about $200 km/h$.
When you hit the ground, the deceleration is always going to be a lot more than $1g$, as the ground tends not to give way. From small heights, your muscles can take the strain; for higher speeds your bones will snap, and at very high speeds, you will not only break lots of bones, but your brain will decelerate so fast that it will stop functioning. That will kill you instantly.
From 10 floors up (say $30m$) you will hit the ground after $1.7 s$, at a speed of $9.8*1.7=17 m/s$ or $60 km/h$. Try driving into a wall at that speed. In a modern car you may survive it but, without the seatbelt, airbags and lots of steel around you, you are almost certainly dead.
To recapitulate, it's the deceleration that hurts you, and that becomes larger as your speeds gets higher and the ground does not start to give until you reach speeds much higher than what will kill you.
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$\begingroup$ thats sort of what i was trying to say with s=ut+(1/2)gt^2 which could be rearranged to get g=(2(s-ut))/t^2 where s=distance, t=time, g=acceleration, u=initial velocity $\endgroup$– ziggyMay 9, 2015 at 10:29
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$\begingroup$ I think the key point (to me) is, it's the negative acceleration (deceleration) that kills you. Thanks. $\endgroup$ May 9, 2015 at 11:03
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Let me summarize the discussion in the comment section. First of all you can safely use Newtons equation of motion
$$ F = m g $$
with a constant gravitational acceleration $g$. This is simply because the height of a 10 stories building, lets say $30 m$, is still very very small compared to the radius of the earth which is about $6000 km$.
Therefore, as you correctly say, the acceleration during the fall is constant and as a consequence the speed $v$ increases linearly with the duration $t$ of your fall
$$ v = g\ t$$
If you are familiar with the concept of energy: the potential energy of your initial position is converted to kinetic energy during your fall and when you want to stop again you have to get rid of this energy.
When you hit the ground there are different possibilities for the kinetic energy to go away. For example the deformation of the ground you are landing on (it surely makes a difference if you land on concrete or on a trampoline), but also deformation of you (that's what hurts).
If you now fall from a greater height you will have more kinetic energy at the moment of your impact and you and the ground will deform more. Unfortunately, the ground will usually deform much less then you do and it will hurt you more.
answered May 9, 2015 at 10:33
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F=ma is saying that if a mass 'm kg' is accelerated at 'a metres per second per second' then there has to be a constant force of 'F' pushing it.
a better equation to represent the effects of gravity would be:
F=(G*m1*m2)/d^2
where: G= universal gravitational constant (6.6738410^-11 m^3 kg^(-1) s^(-2)), g= acceleration due to gravity (9.81 m s^(-1) s^(-1)), m1= mass of object (kg), m2= mass of planet (kg), d= distance between the centre of mass of both objects (m).
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$\begingroup$ How does that answer the question at all? $\endgroup$ May 9, 2015 at 9:46
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$\begingroup$ And besides. The two formulas give in very good approximation the same results to account for a fall from 10 stories $\endgroup$ May 9, 2015 at 9:51
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$\begingroup$ true, its the rapid acceleration (deceleration) that hurts you not the force. In that case, rearrange F=ma to a=F/m sub in the above equation for force to get a=((G*m1*m2)/d^2)/m and you can see why it hurts more when the distance is greater because the acceleration (deceleration) is greater. $\endgroup$– ziggyMay 9, 2015 at 9:53
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$\begingroup$ Actually according to your formula acceleration is smaller when you farer away :P $\endgroup$ May 9, 2015 at 9:54
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$\begingroup$ well if smaller you mean negative, then yes of course deceleration will be less because you would have a high acceleration in the negative direction and hence the value would be negative. $\endgroup$– ziggyMay 9, 2015 at 9:57