How to find initial velocity of projectile?

[Millan]

I've been trying to figure this problem out for a while. Please help.  An object (9kg) is launched off a building (442m height) and lands 500m away from base of building. At what velocity was the object launched?

[Melanie]

How was it launched? Horizontally (angle of zero degrees)? or at some angle?

[Hungry!]

Answer, I think --> 52.7 m/s

I'm guessing that this is a horizontal launch, where ? = 0° ?  If not, I can't answer this problem, and the answer I gave is meaningless.

Given:

m = 9 kg
H = 442 m
R = 500 m
g = 9.81 m/s^2 (Earth gravity, mean)

Mass doesn't matter unless you're including air resistance, which I don't think you are.

Vo = R * SQRT { g / 2H }

Vo = (500 m) * SQRT { (9.81 m/s^2) / [ 2 * (442 m) ] }
Vo = (500 m) * SQRT { (9.81 m/s^2) / [ 884 m ] }
Vo = (500 m) * SQRT { 0.0111 /s^2 }
Vo = (500 m) * (0.105 /s)
Vo = 52.7 m/s

[Jones]

If you want to solve it without jumping to a formula, you can do this:
Find the time that it would take for the object to fall from the top of the building. I'm assuming it was released horizontally, so initial vertical velocity is zero. So,
y2 = y1+Vo t+1/2 g t^2
V0 is 0 and y2-y1 is the height H (just the difference in y values).
H=1/2gt^2
t = sqrt (2H/g)

Since the object was released horizontally, there is no acceleration in that direction.
V=d/t = d/sqrt(2H/g)
=500/sqrt(2*442/9.
=52.6 m/s