Projectile Automobile

## Question:

I was recently given a problem on an exam wherein a car fails to
make a turn and goes off a cliff as a result. The question then
asked what its position was three seconds after going into
projectile motion. We were given its initial velocity as 40m/s
and that gravity was normal. How does one go about finding
position with this information? Thank you for your time.
## Answer:

Cars going off cliffs behave identically with balls thrown from
windows. With an initial velocity of 40 m/s in the horizontal
direction, in 3 seconds it horizontal location will be 3*40=120 meters from the edge of the cliff.
Its vertical position will be below the cliff edge by the
amount which any object falls in 3 seconds under the influence of
gravity.

Gravity near the surface of the Earth accelerates objects
downward at 9.8 meters per second per second. If you remember the
formula for how position depends on time under constant
acceleration you can use that. It is
d=1.2*a*t^{2}. If t=3 seconds, then t^{2}=3*3=9, so d=1/2*(9.8)*9 or d=44.1
meters below the cliff edge.

If you do not remember the formula, you can get it by
recognizing that for a constant acceleration, the average
velocity may be assumed for the whole time interval. The average
is (start+end)/2. Starting downward velocity was zero so average
is just end/2. With constant acceleration the ending velocity is
just the acceleration times the time, or in this case 3*9.8. This
makes the average velocity in the downward direction 3*9.8/2. To
get the final downward position, just multiply the average
velocity by the time of 3 seconds, again getting 44.1 meters
below the cliff edge.

This information is brought to you by M. Casco Associates, a
company dedicated to helping humankind reach the stars through
understanding how the universe works. My name is James D. Jones.
If I can be of more help, please let me know.

JDJ