For the previous example of the out of control bus which is forced to jump from a location $\langle 0,40,-5 \rangle$ with an initial velocity of $\langle 80,7,-5 \rangle$. We have now found the time of flight to be 9.59s and now want to find where the bus returns to the ground?

From the previous problem you already know the final location of the ball in the y direction to be 0 as it has met the ground after 9.59s.

Now to find the range in the x and z directions:

$$ x_f = x_i + V_{avg,x} \Delta{t}$$

$$ = 0 + 80m/s(9.59s)$$

$$ = 767m$$

$$ z_f = z_i + V_{avg,z} \Delta{t}$$

$$ = -5 + -5m/s(9.59s)$$

$$ = -52.95$$

Final position = $\langle 767,0,-52.95 \rangle$ m