Question

A ferris wheel with a radius of $8.0m$ makes $1$ revolution every $10s$.When a passenger 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. $0$
• B. $1.0$m
• C. $8.0m$
• D. $9.1m$

Answer 1

Answer: (D)

$9.1m$

The beginning O of plane coordinate system ${O}_{xy} is located on the ground under the point of release.

$O_y$ axis is directed upward, $O_x$ axis has horizontal direction in the plane of the Ferris wheel.

In such coordinate system ball has coordinates:

$x\left(t\right)=\omega rt,$-------------(1)

$y\left(t\right)=2R-\frac{g{t}^2}{2}$-------------------(2)

here $\omega =\frac{2\pi }{T}$ and $T=10s\approx 0.63rad/s$

R is the radius of the Ferris wheel, $g=9,81m/s^2$

the gravitational acceleration,

t –time after release. If the ball fallen,

then y = 0.

also

$t=2\sqrt{\frac{R}{g}}$-------------------(3)

Putting 3 in 1 we get,

$x=2\omega R\sqrt{\frac{R}{g}}\approx 2\times 0.63\times 8\sqrt{\frac{8}{9.81}}\approx 9.1$