## Calculus 10th Edition

$153.664$ meters
$s(t)=-4.9t^2+v_0t+s_0$ 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