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My confusion arises with free falling body.

  1. For a free falling body the displacement ~ time graph has a kink (at the time when the body hit the ground ). at a kink point, a function is not derivable by the rule of calculus. but we see in the free falling case the body has velocity but in opposite direction at the moment it hit the ground.

    1. For same free falling body as the velocity is a discontinuous function of time (at the time when it hit the ground) there should not be any acceleration because a derivative function must be continuous by the theory of calculus. But velocity $v$ is not continuous at that moment of time (when it hits the ground). But it has an acceleration spike value. So I'm confused very much with this mismatch with mathematical theorem and the practical application in physics. what is the solution??

acceleration and velocity

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    $\begingroup$ acceleration is constant, accept at kink, with value 10m/s^2. Look at last graph. $\endgroup$ – Anubhav Goel Mar 15 '16 at 5:36
  • $\begingroup$ We really don't have any definate direction when body hits ground. Direction is not upwards. $\endgroup$ – Anubhav Goel Mar 15 '16 at 5:39
  • $\begingroup$ then does not it disobey the theorem of calculus.???? $\endgroup$ – pritam Mar 15 '16 at 5:40
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    $\begingroup$ These functions just look broken and discontinuous at normal scale. If you enlarged the time scale so that you can distinguish milli- or microseconds, then you would see that they are all continuous and smooth. $\endgroup$ – Tofi Mar 15 '16 at 14:31
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    $\begingroup$ if we define the continuity or discontinuity on the basis of scale then all function is continuous.....but we don't do that ..isn't it?? just for example in quantum mechanics we talk about Schrodinger eq and after solving the eq we granted the solution on the basis of continuous and discontinuous right?? the continuous is acceptable and discontinuous is not. but if we introduce SCALE concept then every wave function is continuous and acceptable. should we do it ?? the ans is no.. $\endgroup$ – pritam Mar 15 '16 at 15:10
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When the ball makes contact with the ground, the ground exerts a very large (upward) force on the ball for a very short interval of time. This large force causes the ball velocity to change direction from downward to upward, and translates into a large upward acceleration of very short duration. So there is no inconsistency with either the laws of physics or the laws of mathematics.

If the ground is rigid, once the ball makes contact with the ground, the leading edge of the ball comes to a full stop, but the remainder of the ball is still moving downward. The ground exerts a force on the ball, and the ball begins to compress. A compression wave travels upward through the ball. The portion of the ball within the compression zone is not longer moving, but the part of the ball beyond the compression zone is still moving downward. Eventaully, the compression zone encompasses the entire ball, and the entire ball has come to a stop. Next, the compression begins to release. First the part of the ball at the top decompresses, and the velocity of this material is then upward. The decompression wave then travels downward until the ball is fully decompressed, and the entire ball is now traveling upwards. At this point, the ball loses contact with the ground. All these events take place within a tiny fraction of a second.

This description is qualitative, but it captures the essential mechanistic features of what is happening.

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Acceleration of an object, e.g. a free falling ball on earth, is $g=9.81 \frac{m}{s}$. This is a constant, as can be seen in your diagram. Look at the scale - yes acceleration is not zero.

Having this in mind we look at the moment before and after hitting ground. This way we omit looking of the elasticity of the ball. Your diagrams show a perfect elastic ball without dissipative forces. Before ground it has velocity $v$ and a time slice $ \delta t$ later it has veloctiy $-v$.

$$\text{for }\delta t\rightarrow 0\,s\quad a=\frac{v-(-v)}{\delta t} \rightarrow\infty\,\frac{m}{s^2}$$ Because of deformation and internal forces in the ball $\delta t\neq0\,s$. This deformation time is the reason, that acceleration is very large in your diagram, but not infinite.

However for reality e.g. $\delta t=$ and $v=10\,\frac{m}{s^2}$

$$a_{bounce}=\frac{v-(-v)}{\delta t} = 1\cdot 10^4 \,\frac{m}{s^2}\Rightarrow t_{bounce}\approx2\mu s$$

Extending this comprehensible example to the velocity also pictures the ball at the ground. At one time an incompressible ball stops at the ground. I doubt that objects like that exist. A very small time velocity of ball mass center should be minimal during deformation on ground.

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  • $\begingroup$ yes i understand this.but there still i don't get the answer that according to mathematics at the time it hit the ground then at that instant the velocity must not exist as the displacement time graph is not derivable at that point. but we get a non zero velocity this shows velocity exist.how?? moreover if consider the inelastic case then also velocity exist but lesser magnitude because of dissipation . my confusion is there should not exist velocity at that time according to mathematics but it exist. how??? $\endgroup$ – pritam Mar 15 '16 at 15:22
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    $\begingroup$ In reality, the displacement and velocity graphs are continuous and differentiable. Velocity cannot change instantaneously, there is always a period of acceleration. You're looking at an ideal, instantaneous rebound, which requires infinite acceleration - that's where you're finding the disconnect with reality. $\endgroup$ – Nuclear Wang Mar 15 '16 at 15:32

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