# Motion of free fall [duplicate]

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?

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. 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.