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Feynman wrote in his lecture

From Newton’s Second Law we can easily tell how the object moves, and it is easy to find out how the velocity varies with time, namely, that it increases proportionally with the time, and that the height varies as the square of the time. So if we measure the height from a zero point where the object is stationary, it is no miracle that the height turns out to be equal to the square of the velocity times a number of constants

I am not able to understand what he means by saying that "height turns out to be equal to the square of the velocity....."

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  • $\begingroup$ If $v \propto t$ and $h \propto t^2$ then $t \propto v$ and $h \propto v^2$ $\endgroup$ – nluigi Aug 7 '19 at 11:54
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I am not able to understand what he means by saying that "height turns out to be equal to the square of the velocity....."

Here Feynman talks about a freely falling object which was stationary at some point in time

So using the $3^{\text{rd}}$ law of motion we can see what he means. $$v^2-u^2=2as$$ Rearranging, we have $$s=\dfrac{v^2-u^2}{2a}$$

But as the object initially starts from rest the equation reduces to

$$s=v^2\times\dfrac1{2a}$$ where $\dfrac 1{2a}$ is a constant.

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