We know that according to law of free falls object, all bodies fall with the same constant acceleration. But in distance formula ($s = \frac12 gt^2$), why the acceleration is just half?
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$\begingroup$ Possible duplicates: physics.stackexchange.com/q/89590/2451 and links therein. $\endgroup$– Qmechanic ♦May 31, 2022 at 10:29
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2 Answers
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The straight line graph of velocity against time for constant acceleration from zero velocity is shown below.
The area under such a graph (a triangle) is the displacement, $s = \frac 12 \cdot gt\cdot t = \frac 12 gt^2$.
Note that the average velocity is $\frac{v-0}{2}$ and so the displacement is $s=\frac{v-0}{2}\cdot t = \frac {gt}{2} \cdot t = \frac 12 gt^2$ as before.
If there is an initial velocity $u$ then the displacement (tapezium or triangle plus rectangle) is $s = ut+\frac 12gt^2$ which can be found using the graph below.
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The acceleration is not half of $g$. It is equal to $g$. The expression you are referring to is obtained by integration. \begin{align} &\frac{d^2s}{dt^2} = g \\ \implies &\frac{ds}{dt} = gt+c_1 \\ \implies &s = \frac{1}{2}gt^2+c_1 t+c_2 \end{align}
If you take the initial conditions as $s(0) = 0$ and $s'(0) = 0$, you will get the expression.
