A water balloon is thrown at a 30 degree angle above the horizontal from a balcony

The balloon remains in the air for 2.5 seconds before landing 43.3 meters from the base of the balcony. How fast was the water balloon traveling at the apex of its trajectory? How fast was the water balloon thrown from the balcony?
Thanks for your help! :D

The balloon is thrown at an angle of 30º above the horizontal. Normally we'd go ahead and resolve that vector into its horizontal and vertical components, but since we don't KNOW the launch velocity, we can only do a partial job:

vx (horiz. velocity) = v*cos(30º)

vy (vert. velocity) = v*sin(30º)

That's the best we can do for now. We'll come back to it later.

The balloon traveled horizontally for 43.3 meters before splashing down, and was in the air for 2.5 seconds. Since we know the horizontal velocity should be constant (with no forces acting horizontally to provide acceleration), we can easily figure out the horizontal velocity now:

vx = Δdx / t

vx = (43.3 m) / (2.5 s)

vx = 17.32 m/s

Incidentally, this is also the balloon's speed at the apex of its trajectory. At the apex, the balloon has NO vertical velocity, so all of its velocity must be horizontal. Since the horizontal velocity stays at 17.32 m/s for the entire trip, that's also its apex velocity.

Now...let's go back to the launch velocity. We learned earlier that:

vx = v*cos(30º)

And now we know that vx = 17.32 m/s, so...

17.32 m/s = v*cos(30º)

v = (17.32 m/s) / cos(30º)

v = 20 m/s

I hope that helps. Good luck!

43.3 / 2.5 = 17.32 m/s: Horiz. component
17.32 / COS 30 = 20 m/s: thrown from balcony @ 30 Deg
SQRT (20^2 - 17.32^2) = 10 m/s Vert component

Get a stop watch and experiment. Hope this helps.