A ball is dropped from a platform 19.6 m high. Its position function is 

We have $a=\frac{d^{2}x}{d{t}^2}=-9.8$. The initial conditions are  $x \left(0\right) = 19.6$ and $v \left(0\right) = 0.$

So     $v=\frac{dx}{dt}=-9.8t+v\left(0\right)=-9.8t$

$\therefore$   $x=-4.9t^2+x\left(0\right)=-4.9t^2+19.6$

Now, the domain of the function is restricted since the ball hits the ground after a certain time. To find this time we set $x = 0$ and solve for $t$.

$0=-4.9t^2+19.6 \Rightarrow t=2$

Thus $x=-4.9t^2+19.6 (0\le t\le 2)$