Fairgoers ride a Ferris wheel with a radius of 6.00 m, the bottom of which is 1.75 m from the ground. The wheel completes one revolution every 28.0 s. A passenger drops his keys when he is three quarters of the way up and at an angle of 45° to the vertical. Where do the keys land relative to the base of the ride? (Take the positive direction to be to the right.)
x = (Vo cos theta)t
Vx = Vo cos theta
Vy = Vo sin theta - gt
Vox = Vo cos theta
Voy = Vo sin theta
Vavg = (2)(Pi)(R/T)
The Attempt at a Solution
I first calculated Vo (which, I assume, is constant for a ferris wheel, so Vo = V or Vavg). I really didn't know where to go after that from the above equations, since to calculate x, the distance, we need time. To calculate time, we need Vy. I attempted to calculate initial velocity (Voy = Vo sin theta), which was .95m/s and plug it into the 3rd equation as Vy(as velocity should be constant), to get time... then plug that into the first equation to get the distance... but that didn't seem to work.
I've tried different formulas and manipulating many things to try and get the answer, but I seem to be stuck. I'm guessing it's something do to with the height and free fall to get t maybe, since I can't seem to get a true t with any of the equations of projectile motion.
Any assistance would be greatly appreciated. Thanks.