#### Answer

$153.664$ meters

#### Work Step by Step

$s(t)=-4.9t^2+v_0t+s_0$
When the rock is dropped, it implies that the initial velocity is $0$ meters per second.
$v_0=0$
Let the ground level be $s=0$. The rock hits the ground at time $t=5.6$. Therefore,
$s(5.6)=0$.
$s_0$ would be the height when the rock is dropped, which is also the height of the building.
$s(t)=-4.9t^2+s_0$
$s(5.6)=-4.9(5.6)^2+s_0$
$0=-4.9(5.6)^2+s_0$
$4.9(5.6)^2=s_0$
$s_0=153.664$ meters