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The Q is:

enter image description here

My teacher said that the acceleration of the ball during the contact will be different from 'g' and therefore we have to treat the downward motion and upward motion separately .

My question is how and why the acceleration of the ball is different from 'g' during the contact ? As at the time of contact also the accelaration will be acting downwards only . Please clarify my doubt .

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Remember that he acceleration is the result of the sum of all forces. when the ball is in the air the only force is gravity (if we neglect air friction). When it is touching the floor, the floor is making a normal force on the ball, in the same way that there is a normal force when you have a block resting in a surface. The normal force prevents the ball from going into the floor, and is strong enough to actually make it bounce back.

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  • $\begingroup$ thanks a lot, sir ... I clearly understand now $\endgroup$ – Garima Singh Sep 15 at 7:08
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During contact there is a force applied to the ball. This contact time is very short and is often modeled as an impact. But regardless, when the bat hits the ball it applies a force and hence an acceleration given by F=ma. Gravity applies a force downward while the bat applies a force dependent on the strike angle. The accelerations are in the same direction as the forces. These are independent and final acceleration is vector addition of the 2.

But once the ball loses contact with the bat the only force remaining is due to gravity. So for the time the ball and bat are in contact, there are two forces on the ball and hence two vector accelerations.

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  • $\begingroup$ Not sure what was there before editing, but there's no mention of a bat in the question as it currently stands. $\endgroup$ – The Photon Sep 15 at 5:07
  • $\begingroup$ @jmh: thanks a lot, sir ... although there was no mentioning of the bat but your explanation is like an example to my query. $\endgroup$ – Garima Singh Sep 15 at 7:10
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The graph of velocity against time which which might have been expected is shown on the left.

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Whilst in the air the acceleration of the ball is $g=+9.8\,\rm m\,s^{-2}$ and this is the gradient of the velocity time graph.

There is a problem with this graph in that at time $2\,\rm s$ the gradient of the graph is infinite which implies that the acceleration is infinite.
There cannot be an infinite acceleration as this would imply that an infinite force is acting on the ball when it is in contact with the ground.

Expanding the time axis in the region of the red rectangle the graph might look like that on the right.
Here it was assumed that the acceleration of the ball when in contact with the ground for $\frac{2}{1000}^{\rm th}$ of a second was constant and its value can be worked out as being $-19,600\,\rm m\,s^{-2}$.
This means that on the time scale of the left graph the line at time $2$ seconds is effectively vertical.

Note that the graph of the right still cannot be correct as, as drawn, the gradient is not defined at times $1.999\,\rm s$ and $2.001\,\rm s$.
So what might the graph actually look like around those times?

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  • $\begingroup$ thanks a lot ... great explanation as always. since in the question it has asked to neglect the small interval during which the ball is in contact with the ground ... I think in that case the graph at left is appropriate. $\endgroup$ – Garima Singh Sep 15 at 8:02

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