A Ferris wheel with a radius of 8.0m makes 1 revolution every 10 s.When he is at the top, essentially a diameter above the ground, he releases a ball.How far from the point on the ground directly under the release point does the ball land?
A)0 m
B)1.0 m
C)8.0 m
D)9.1 m
E)16 m

Question

Quizplus · Accepted Answer

The initial velocity of the ball is horizontal because it is released at the top of the Ferris wheel. The distance it travels horizontally is the circumference of the wheel, which is 2π(8.0m) = 50.3m. The time it takes for the ball to hit the ground is half of the period of the Ferris wheel, which is 5s. Using the equation d = vt + 0.5at^2 with vertical initial velocity of 0 and acceleration due to gravity of 9.81 m/s^2, we get:
d = 0 + 0.5(9.81 m/s^2)(5 s)^2 = 122.6 m
So the ball lands 122.6m away horizontally from the release point, and we can use the Pythagorean theorem to find the distance from the point on the ground directly under the release point:
√(8.0 m)^2 + (122.6 m)^2 ≈ 9.1 m

A Ferris Wheel with a Radius of 8