Question

A Ferris wheel with a radius of 8.0 m makes 1 revolution in 10 s. When a passenger is at the top, he releases a ball. How far from the point on the ground directly under the release point does the ball land?

Answer 1

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:

$x(t)=ωrt~~~~~~~-(1)$

$y(t)=2R−\dfrac{gt^2}{2}~~~~-(2)$

here $ω=\dfrac{2\pi}{T}$ and $T=10s=0.63rads^{-1}$

R is the radius of the Ferris wheel, $g=9.81m/s^2$

 

 

Let t –time after release. If the ball fallen, 

then y = 0. 

also 

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

Putting 3 in 1 we get,

$x=2ωR\sqrt{\dfrac{R}{g}}=2\times 0.63\times 8\sqrt{\dfrac{8}{9.81}}=9.1m$