Question

A Ferris wheel 50ft in diameter makes one revolution every 40sec. If the center of the wheel is 30ft above the ground, how long after reaching the low point is a rider 50ft above the ground?

Answer 1

Answer:

The rider is 15.91  seconds  50 ft above the ground after reaching the low point

Step-by-step explanation:

We can evaluate the angle  α

using trigonometry applied to the orange small triangle with height 50-30 = 20 ft and  hypotenuse equal to the radius  r = 25ft

Now

$20 = \frac{25}{sin(\alpha)}$

$\alpha = arcsin{\frac{20}{25}$

$\alpha = 53.13^{\circ}$

So  50 ft  of height corresponds to the total angle:

$90^{\circ} =53.13^{\circ} = 143.13 6^{\circ}$  = 2.498 radians

Now the angular velocity

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2 \pi}{40}$

$\omega =0.157rad/s$

To describe  2.498  rad  it will take:

$t = \frac{2.498}{0.157}$

t = 15.91 s