A question regarding equations of motionSuppose I drop a body from a height h. The initial velocity is zero. Now it reaches the ground and stops and finally becomes zero. So now u=0 and v=0. If I use the equation of motion v = u + a t to find value of acceleration and input u=0 and v=0 I get a=0 even though it is clearly wrong and there was acceleration due to gravity. So what is going on here? Where is my understanding going wrong?

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A question regarding equations of motionSuppose I drop a body from a height   h. The initial velocity is zero. Now it reaches the ground and stops and finally becomes zero. So now u=0 and v=0. If I use the equation of motion   v  =  u  +  a  t to find value of acceleration and input u=0 and v=0 I get a=0 even though it is clearly wrong and there was acceleration due to gravity. So what is going on here? Where is my understanding going wrong?

iletsa2ym · Accepted Answer

The equations of motion only work for constant acceleration. The setup mentioned in your question doesn’t satisfy that criteria. You can use the equations of motion to model the motion of the body right from the instant of time you drop it, to the instant of time right before it hits the ground/zero level. In this period of time, the acceleration is constant (almost constant) and the equations of motion hold. However, after hitting the ground, energy is lost to heat, friction, sound, and other forms of energy transfer take place. Also, the ground exerts a force on the body. All this is to say that the value of acceleration is no longer constant, and the equations of motion no longer hold.

Cael Dickerson · Accepted Answer

The body&#039;s motion has two very different parts: It begins by accelerating down, until just before reaching the ground. At this point, as you noted, its velocity will be   v  =  −  g  ∗  t (if the acceleration due to the earth gravitation is   −  g) and you can calculate this   t using the equation for the displacement in your case (with   x  =  h,       v    0    =  0)   −  h  =  1      /    2  g      t    2  .So far the equations will fit very well.But, when the body hits the ground at the calculated speed, as you know it will have to stop... An elastic body will actually jump up and down a few times, losing a bit of the energy to heat. An inelastic body will lose all the energy almost at once to heat. Eventually, perhaps after a couple of small jumps, the body will settle down on the ground with velocity 0. But you can&#039;t use the equation for free fall at a constant acceleration to calculate that.

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