An airplane is flying horizontally with a speed of 1000km/h (280 m/s)
when an engine falls off.  Ignoring air resistance, if it takes 10s
for the engine to hit the ground
a) how high is the plane?
b) how far horizontally does the engine travel while it falls?
c) if the airplane somehow continues to fly as if nothing happened,
where is the engine relative to the airplane at the moment the engine
hits the ground?

to solve for height use 
this is actually relatively simple to calculate using the equation
s=VoT+1/2aT^2
where s is the distance and Vo is intial velocity, T is time and a is
acelleration due to gravity
plug in Om/s for Vo because it has no velocity in a downward path and
then 10seconds for time, then use 9.8m/s^2 for a and solve for s
to solve for position
the plane is directly over the position where the engine strikes the
ground at 10 seconds because both have the same horizontal velocity.
and have traveled the same distance because of this.

sorry left off the answer the height is 490m

HW question, right?
To continue from where gregorious left, the remaining answers are:
(b)Horizontal distance = 280 m/s X 10 s
                       = 2800 m.
(c)The engine is at a diagonal distance of 2842.55 m (sq root(2800^2 + 490^2))  
   behind the aeroplance. The angle between the the horizontal and the
straight line joining the engine and aeroplane is 9.93 degrees (arc
tan (490/2800)).

Sorry, need to change answer (c) as I've calculated wrongly and
Gregorious answer is correct.
The correct answer for (c) is
(c)the engine is 490 m below the airplane at the moment the engine hits the ground.

Good luck to your HW!