Projectile Motion Problem

The y-graph represents the problem, because the car begins at above the origen then is thrown to the Earth after a certain amount of time goes by.  The slope is always negative, so the derivative and Vy graph is a negative linear line.  As time goes by, the position approaches 0.  As it approaches 0, it approaches 0 faster.  The object starts at rest and begins going faster and faster in a negative direction.

The x-graph represents the problem, because as time goes on the car is still horizontally moving.  The derivative of a positive linear line the poitive position go in a positive direction at a positive rate.

Equations:
First, I used the position equation.  The car travels from 0 meters to 125 meters, so the position is -125 meters.  The vertical motion doesn't have an initial velocity.  Gravity pulls on the car at -9.8 m/s2.  I then solve the equation to find the time the car took to reach the ground.  I find after 5.05 seconds, the tornado replaces the car on the ground.

Next, I use the change in x equation, change in x equals Vx times t, to find how far horizontally the car actually travels.  The tornado flings the car horizontally at an initial velocity of 90.0m/s.  Since I already found the time the car takes to reach the ground, I can plug 5.05 seconds in for time.  I come up with the horizontal displacement of the car by the tornado to be 454.5 meters.