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In a free fall, on impact with the ground, the body experiences a massive force because of a huge acceleration since the velocity goes to 0 rather quickly.

Since this force has to come as a result of reaction, can it be assumed that the body applied a massive force on impact far greater than its weight?

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  • $\begingroup$ The reaction is going to be at a minimum equal to the weight because if not you would accelerate downwards. How much MORE is a question that depends on the details. What about falling from low height onto a trampoline, vs a marble floor. $\endgroup$ Commented Oct 25, 2013 at 19:07
  • $\begingroup$ Through a separate thread, I understood that falling on a trampoline or cotton is somewhat like the case of compressing a spring. The impact force is 0 but keeps increasing till it becomes equal to the weight of the object and the body stops and a motion in the reverse direction could take place as the base object tries to regain its shape. $\endgroup$
    – pran
    Commented Oct 25, 2013 at 19:25
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    $\begingroup$ Actually the impact force will increase beyond the weight. The extra force is used to accelerate the object away from the contact. If at peak deflection the forces where equal then the system would remain in equilibrium and nothing would move from that point on. $\endgroup$ Commented Oct 25, 2013 at 19:45

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You are right. In the case of a sudden stop a large force is exerted on the falling body to bring its velocity to $0~m/s$ in a short amount of time. A useful quantity related to force that describes this is the "impulse", which is defined as the integral of force with respect to time:

$ \begin{equation} J=\int_{t_1}^{t_2}{F\cdot dt} \end{equation} $

where $J$ is the impulse. This quantity is also equal to the change in momentum of an object. In the case of our free falling object, assuming the object starts at the top of its fall at rest relative to the ground, the change in momentum from the beginning of the fall to right before impact has the same magnitude (but is opposite in direction) as the change in momentum during the impact; thus, $|J|$ is the same for both the time the object is falling and the time the object is falling. In the case of free fall near the earth's surface, $F$ is constant and the integral above evaluates to $F\cdot \Delta t$. Solving for $F$ and taking the magnitude of both $F$ and $J$ we find

$\begin{equation} |F|=|J|/\Delta t . \end{equation} $

Since the stopping time is much smaller than the falling time, and $|J|$ is the same for both falling and stopping, we find that the force of impact on the ground is much greater than the force during free fall (since dividing by a smaller $\Delta t$ will give a larger $|F|$. Since the force experienced during free fall is just the force of gravity, $mg$ (or simply the weight of the object) it is true that

the body [experienced] a massive force on impact far greater than its weight

as you stated in your question.

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