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Simon777
Simon777 is offline
#5
Sep7-11, 09:59 AM
P: 35
Quote Quote by ehild View Post
You have the formulae

x=v0xt
y=y0+v0yt-g/2 t2.

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 v0y?
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