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I have shown (using a lot of assumptions made at the high school level of physics) that if a ball is thrown upwards or downwards, as long as the initial speed is the same, the final speed at the moment the ball hits the ground is also the same.

Despite knowing how the math works, I would like some intuition behind why this is true. To me, there is no reason why they would be the same. Does this have to do with the conservation or energy or something?

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  • $\begingroup$ I have attempted to make the title easier to understand, and to fix the confusion that @Qmechanic pointed out. Please confirm I did not accidentally change the question to something you didn't intend to ask. $\endgroup$ – Floris Feb 8 '17 at 23:18
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I assume that the assumptions that you made are essentially that there is no loss of energy. You say you did the math already, so I will try to give you an "intuitive" method of understanding it, although I think the math shows it more clearly.

Basically, your ball starts with a total amount of energy. In this case, it can only take two forms : kinetic energy, $\frac{1}{2}mv^2$ where m and v are respectively the mass and speed of the ball, and gravitational energy, $mgh$ where h is the height at which the ball is located (height relative to an arbitrary point).

So, if you start at the same height, AND at the same speed, two balls thrown will always have exactly the same energy capital. And since we assumed that there is no dissipation of energy, this energy capital is constant.

So, any balls thrown at the same height AND speed, no matter the angle, will have the same energy. This also means that, if we measure their speed at a later time, it will only depend on their initial energy, and on their current height. More specifically, when your two balls hit the ground, they are at the same height (say $h=0$) and they also have the same energy (since the only difference on their throws was the angle), hence, they have the same speed.

To resume in a few equations, if $E_0$ is the initial energy of the ball (which depends only on its initial speed $v_0$ and height $h_0$), then the speed of the ball at any given height is :

$$v = \sqrt{\frac{2(E_0-mgh)}{m}}$$

As we can see, it does not depend on the angle with which the ball is thrown.

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As you mentioned in the question, some simplifying assumptions are made in physics. The key assumption will be that there is no air resistance. The ball loses no energy* (and gains no energy) when it is in the air, so if it landed with a speed any different to the one it landed with, we would be faced with the confounding prospect of having to explain where the difference in energy came from. The conservation of energy is not a simplifying assumption - it is a universal law, so we're in trouble if something ends up randomly changing its energy.

In real life, if you could measure a ball's speed at projection and its speed on impact, you would find that these generally are a little different, because air resistance is a real phenomenon. Physics improves with time and experimentation, and Stokes (1851) devised a law to improve on the thousand-year old equations of motion which assumed no drag. But, on a clear day with low wind, and for fairly short distances, the equations predict that a ball will land at the same speed it is projected, and this turns out to be fairly accurate.


*Mechanical energy falls into two types: kinetic and gravitational potential. Throughout the motion, the ball is converting K.E. into G.P.E. and then back again, but never actually loses any overall energy.

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