# 1 d motion free fall problem

1. Dec 29, 2012

### baird.lindsay

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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...?

2. Dec 29, 2012

### cosmic dust

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.

3. Dec 29, 2012

### Staff: Mentor

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.

4. Dec 29, 2012

### baird.lindsay

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...

5. Dec 29, 2012

### Staff: Mentor

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?

6. Dec 29, 2012

### baird.lindsay

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

average speed is 33/1.2 =27.5 m/s?? sorry I'm confused ...:(

7. Dec 29, 2012

### Staff: Mentor

No. 33 m is the distance the car falls during its free fall. But it hasn't reached the ground yet.
Good.
How fast is it going when it begins to slow down? (See the answer to part a.)

8. Dec 29, 2012

### cosmic dust

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...

9. Dec 29, 2012

### baird.lindsay

Part D)

Last edited: Dec 29, 2012
10. Dec 29, 2012

### baird.lindsay

okay .5 X 21.255 (acceleration part b) X 1.2^2 =15.516 meters plus 33=48 meters?

11. Dec 29, 2012

### Staff: Mentor

That's the idea (but be careful about rounding off too soon).

12. Dec 29, 2012

### baird.lindsay

thank you all figured it out!!