Where Does a Stuffed Animal Land When Dropped from a Ferris Wheel?

[wadini]

Fairgoers ride a Ferris wheel with a radius of 5.00 m . The wheel completes one revolution every 31.0 s
If a rider accidentally drops a stuffed animal at the top of the wheel, where does it land relative to the base of the ride? (Note: The bottom of the wheel is 1.75 m above the ground.)

okay so I got that velocity which is 1.01 m/s of the rider on the ferris wheel but I really don't know where to go from there...any suggestions? It seems simple but I am stumped!

 

[djeitnstine]

Find the time taken for the stuffed animal to reach the ground. Use this time to find where the base will be when the object has landed.

 

[nlightner]

I would approach this question by using the principles of physics to determine the answer. First, we need to understand the motion of the stuffed animal as it falls from the top of the Ferris wheel. We know that the stuffed animal has an initial velocity of 1.01 m/s, which is the same as the rider's velocity at the top of the wheel. We also know that the stuffed animal will experience a constant acceleration due to gravity of 9.8 m/s^2.

Using the equation d = v0t + 1/2at^2, where d is the distance traveled, v0 is the initial velocity, a is the acceleration, and t is the time, we can calculate the distance the stuffed animal will fall in 31 seconds (the time it takes for the wheel to complete one revolution). Plugging in our values, we get d = (1.01 m/s)(31 s) + 1/2(9.8 m/s^2)(31 s)^2 = 15.7 m.

Now, we need to consider the height of the bottom of the wheel from the ground, which is 1.75 m. This means that the stuffed animal will land 15.7 m + 1.75 m = 17.45 m from the base of the ride.

In conclusion, if a stuffed animal is dropped from the top of a Ferris wheel with a radius of 5.00 m, it will land approximately 17.45 m from the base of the ride, assuming there is no air resistance or other external factors affecting its motion.