Fairgoers ride a Ferris wheel with a radius of 5.00 {\ m m} . The wheel complete

Question

Fairgoers ride a Ferris wheel with a radius of 5.00 { m m} . The wheel completes one revolution every 31.0 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? (Note: The bottom of the wheel is 1.75 { m m} above the ground.)

Give the answer in meters please!

Fairgoers ride a Ferris wheel with a radius of 5.00 { m m} . The wheel completes one revolution every 31.0 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? (Note: The bottom of the wheel is 1.75 { m m} above the ground.)

Explanation / Answer

angular speed = 2*pi / time of 1 rotaion = 2*pi / 31 = 0.202683397 rad/sec

so.. horizontal speed at the top = angular speed * radius = 0.202683397 * 5 = 1.013416985 m/sec

at the top there wont be any vertical speed ...

so.. along vertical direction ..

intial velocity u = 0

acceleriaon = a = 9.81 m/sec2

distance travelled in vertical direction = s = 10 + 1.75 = 11.75 m

let the time taken to reach the gound be t..

so.. s= ut + 0.5 * a*t^2

so.. 11.75 = 0 + 0.5*9.81*t^2

so.. t= 1.547745 secs ..

so... along horizontal direction ..

speed = vx = 1.013416985 m/sec

time = t = 1.547745 sec

so... as there is no accelertion in horizntal direction ..

distance = vx * t = 1.013416985 * 1.547745 = 1.568511 m

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