Quote by ehild
You have the formulae
x=v_{0x}t
y=y_{0}+v_{0y}tg/2 t^{2}.
You know t, the horizontal distance and the height from where the motorcycle jumps (15 m) with respect to the final height (zero). Can you get v_{0y}?
If you have both the x and y components of the velocity, can you find its angle with the positive x axis?
ehild

I went that route at first and ended up with a wrong answer. I've checked my calculations 5 times and keep coming up with the same wrong angle so if you wouldn't mind double checking me, I'd appreciate it.
y − yo = (voy)t −(1/2)g t^2
15m = (voy) (3s)  (1/2) (9.81m/s^2)(3s)^2
59.145 = (voy) (3s)
voy = 19.7 m/s
x − xo = (vox)t
69.4m = (vox)(3s)
23.13 m/s = vox
To find the angle I used:
tan^1 of (19.715/23.13) = 40.44 degrees