Calculating Free Fall Motion on a Drop Ride: Solving for Velocity and Distance

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SUMMARY

The discussion focuses on calculating free fall motion for a drop ride, addressing key physics equations and concepts. The final velocity of passengers after 2.6 seconds of free fall is -25.506 m/s, while the average deceleration during the stopping phase is +21.255 m/s². The total distance traveled during the ride is confirmed to be 66 meters, with additional calculations for a shoe dropped from a height of 33 meters, which reaches the ground at a velocity of approximately 48.4 m/s. The overall time for the shoe to hit the ground is determined to be 3.8 seconds.

PREREQUISITES
  • Understanding of kinematic equations, specifically vf = vi + at
  • Knowledge of gravitational acceleration, g = 9.8 m/s²
  • Familiarity with average speed calculations
  • Ability to manipulate quadratic equations for displacement
NEXT STEPS
  • Learn how to apply kinematic equations in real-world scenarios
  • Study the concept of average speed and its significance in motion analysis
  • Explore the derivation and application of the equation h = 1/2 g Δt²
  • Investigate the effects of air resistance on free fall motion
USEFUL FOR

Physics students, educators, and anyone interested in the mechanics of motion, particularly in the context of amusement park rides and free fall dynamics.

baird.lindsay
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Homework Statement



Passengers on the drop, a free fall ride at the fair, sits in cars that are raised to the top of the tower. the cars are released for 2.6 s of free fall.
a) how fast are the passengers moving at the end of the free fall phase? -25.506 m/s
b) if the cars in which they ride come to rest in 1.2 s what is the avg acceleration? +21.255 m/s^2
c) what is the total distance they ride? 66 meters
d) one girls shoes falls off at the top when the ride begins. how fast is it moving when it reaches the ground?
e) how long is the shoe on the ground when the girl arrives?

Homework Equations



vf=vi+at
s=volt+1/2at^2
vfs^2=vis^2+2as deltax

The Attempt at a Solution

Im stuck on d and e.
I thought vi=0 vf=? xi=33 xf=66 a=-9.8m/s^2
and i tried to use vf=vi+at but i get 25.48 and my notes say it should be 48.4 m...but I might have written my notes wrong...?
 
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Revise c) (Δx = 1/2 g Δt2 and include the displacement during deceleration) I found 48.4 m
About d), what is the height of the tower? Use the formula v = √(2gh)
About e), the fall lasts 2.6s+1.2s = 3.8s. Find shoe's displacement fot that time and subtract it from the height.
 
baird.lindsay said:
Im stuck on d and e.
I thought vi=0 vf=? xi=33 xf=66 a=-9.8m/s^2
What's the height of the shoe when it starts falling? When it hits the ground?

I also agree with cosmic dust: you must redo part c.
 
Revise c) (Δx = 1/2 g Δt2 and include the displacement during deceleration) I found 48.4 m

thank you...can u let me know how to find the displacement during deceleration? I'm not sure what this means...
 
baird.lindsay said:
thank you...can u let me know how to find the displacement during deceleration? I'm not sure what this means...
How far does the car travel while it is slowing down (the 1.2 second portion of the trip)? Hint: What's the average speed during that time?
 
Doc Al said:
What's the height of the shoe when it starts falling? When it hits the ground?

I also agree with cosmic dust: you must redo part c.

The height of the shoe when it stars falling is 33 meters. It hits the ground at 0 meters. Δx=-33

Doc Al said:
How far does the car travel while it is slowing down (the 1.2 second portion of the trip)? Hint: What's the average speed during that time?

average speed is 33/1.2 =27.5 m/s?? sorry I'm confused ...:(
 
baird.lindsay said:
The height of the shoe when it stars falling is 33 meters.
No. 33 m is the distance the car falls during its free fall. But it hasn't reached the ground yet.
It hits the ground at 0 meters.
Good.
average speed is 33/1.2 =27.5 m/s??
How fast is it going when it begins to slow down? (See the answer to part a.)
 
cosmic dust said:
About e), the fall lasts 2.6s+1.2s = 3.8s. Find shoe's displacement fot that time and subtract it from the height.

Sorry, I have misread the question. You should find the time needed for the shoe to touch the ground, solving the equation h = 1/2 g Δt2 for Δt. Thren subtract it from 3.8s...
 
Part D)
 
Last edited:
  • #10
okay .5 X 21.255 (acceleration part b) X 1.2^2 =15.516 meters plus 33=48 meters?
 
  • #11
baird.lindsay said:
okay .5 X 21.255 (acceleration part b) X 1.2^2 =15.516 meters plus 33=48 meters?
That's the idea (but be careful about rounding off too soon).
 
  • #12
thank you all figured it out!
 

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