Given,
The initial horizontal velocity of the car, u=54 m/s
The height from which the car was dropped, h=256 m
The initial vertical velocity of the car will be zero.
Thus from the equation of motion, the height from which the car was dropped can be written as
[tex]h=\frac{1}{2}gt^2[/tex]Where g is the acceleration due to gravity and t is the time it takes for the car to reach the ground.
On substituting the known values,
[tex]\begin{gathered} 256=\frac{1}{2}\times9.8\times t^2 \\ t=\sqrt[]{\frac{256\times2}{9.8}} \\ =7.23\text{ s} \end{gathered}[/tex]The range of flight of car, i.e., the distance between your house and the point where the car lads is given by,
[tex]R=ut[/tex]On substituting the known values,
[tex]\begin{gathered} R=54\times7.23 \\ =390.42\text{ m} \end{gathered}[/tex]thus the car will land at a distance of 390.42 m away from your house.