[dragosani]

I just need to know if I've done this right. If not, please explain where I've gone wrong. Thank you.

A bomber flying in level flight at 200m/s drops a 500kg bomb from an altitude of 3000m.
A.)Find the horizontal and vertical components of the velocity just before it strikes.
B.) How far has it traveled horizontally after release when it strikes.
This is what I have:
Y= 3000m
V=200m/s
m=5ookg
a= 9.8m/s^2

Vx is constant so it would be 200m/s
Vy is zero and just accelerating.

So, using : Y=Yo+ Vot+ .5at^2
3000m= o+0+ .5(9.8)t^2
3000m/4.9m/s^2 = t^2
24.7s=t

Using this time and X=vt I get
X=200m/s(24.7s)
x=4940m

But this just doesn't seem right.

[harum]

You answer is probably correct! Are you confusing the distances travelled in X and Y? Remember though that you have to take into consideration the directions of the initial velocity and the force of gravity, meaning that those things are vectors with both magnitude and direction. Besides, always draw a picture and introduce a convenient frame of reference. There are many ways to select the frame of reference, but once selected all forces have to be checked against that reference frame, in terms of direction. In your particular case, you have selected the ground is Y=0, and initial initial height is Yo=3000, that is the positive direction for the Y axis of your frame of reference is upwards, perpendicular to the ground. This observaton makes the force of gravity, which is always directed to the ground, negative. This means that the acceleration of the bomb along the Y axis is not "a", but "-a" (because the direction is opposite to the positive direction of the Y axis. Your equation should be "Y=Yo+ Vot - .5at^2".