1. The problem statement, all variables and given/known data A mischievous young man riding a ferris wheel decides to release a beanbag as he passes the top point in the ride. The ferris wheel has a radius of 3.39 m and goes around once every 0.95 minutes (or 57 seconds if converted). The bottom of the wheel is 1.03 m off the ground, so that the release point of the bag will be 7.81 m off the ground. 1) What is the direction of the bean bag's acceleration just before it is released by the mischievous young man? 2) What is the speed of the bean bag just before it is released by the mischievous young man? 3) What is the magnitude of the acceleration of the bean bag just before it is released by the mischievous young man? 4) For how much time is the bean bag in the air? 5) The young man finds that the bag does not land directly under the point of release. How far does the bag travel horizontally as it is falling? 2. Relevant equations v=2πR/T a=4π2R/T2 3. The attempt at a solution 1) is the answer up? i thought it was up because he is still rising a little bit until he reaches the very peak of the ferris wheel and releases the bean bag, thus making the direction of the bean bags acceleration up. the other choices were: ....in the direction of motion of the young man's hand ....the acceleration is zero, thus no direction ....opposite to the direction of motion of the young man's hand ....down 2) v=2π(3.39) / 57 = 0.374 m/s i wasnt sure about this one because of the extra 1.03 m above the ground and i wasnt sure if this somehow had to be factored in. the answer just seems really slow but i guess ferris wheels dont move fast haha. 3) a=4π2(3.39)/(57)2 = 0.0412 m/s2 again not sure about this because im not sure if i have to factor in the 1.03 m height somehow 4) Not really sure how to find this one.... 5) i think i need to know the time before i solve this one..... Thanks for any help!