Fairgoers ride a Ferris wheel with a radius of 5.50 m, the bottom of which is 1.75 m above the ground. The wheel completes one revolution every 32.0 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? (Neglect air resistance.)

To determine where the stuffed animal will land relative to the base of the ride, we need to analyze the motion of the object.

First, let's determine the height of the top of the Ferris wheel from the ground level. Given that the bottom of the wheel is 1.75 m above the ground, the total height of the Ferris wheel is the sum of its radius (5.50 m) and the height of the bottom, which is 1.75 m. So, the height from the ground to the top of the wheel is 5.50 m + 1.75 m = 7.25 m.

Next, let's determine the time it takes for the stuffed animal to reach the ground after it is dropped at the top of the wheel. The wheel completes one revolution every 32.0 s, which means it takes 32.0 s to travel 360 degrees or a full circle.

Since the stuffed animal is dropped at the top of the wheel, it will take the same amount of time for it to reach the ground as it takes for the wheel to complete one revolution, which is 32.0 s.

Now, we can calculate the distance the stuffed animal will fall during this time. The distance fallen can be determined using the formula for the vertical distance traveled under free fall acceleration:

d = 1/2 * g * t^2,

where d is the distance fallen, g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds.

Substituting the given values, we have:

d = 1/2 * 9.8 m/s^2 * (32.0 s)^2 = 1/2 * 9.8 m/s^2 * 1024 s^2 ≈ 5065.6 m.

Therefore, the stuffed animal will fall approximately 5065.6 m relative to the base of the ride.