# ZERO Launch angle

1. Jul 2, 2006

### jhrnndz1

I have a problem that I'm having trouble with. The question is, A farris wheel that has a radius of 5 meters, takes 32 seconds to complete one revolution. (a) what is the average speed of the rider on the Ferris Wheel? (b) If passenger drops keys when at top of the wheel, where does it land relative to the base. and finally (c) if passenger drops keys when on way up and is at the 10 0'clock position, where do the keys land relative to the base? (Hint: The bottom of the wheel is 1.75 m above the ground.

Can someone help? Thanks!:)

2. Jul 2, 2006

### DaveC426913

Sure. Show your work.

3. Jul 3, 2006

### jhrnndz1

alright, for part a, i used the zero launch angle equation, (x=vo(t)) and I put x=2pie(r), and put the time as 32 sec, and got .98m/s for the velocity. For part b, i used the equation (x=vo(sqareroot of (2h/g)))and got 1.5 meters. Now, I'm a little stuck on part c and am not quite sure how i should set up the problem. I kind of think that I should be using a general launch angle equation. But i'm not sure, can you help me out on setting up the problem. Thanks.

4. Jul 3, 2006

### vijay123

hey, i calculated your last sum. it is pretty confusing but u ll soon get it. is the ans. by any chance 1.23664m meters?
anyway, the angle at which the key leave the feeris wheel is 30 degres.this is because, the key is trabelling tangential to the circlular ferris, hence it is directly perpendicular to the radius.now 10 0 clock position is actually 60degress from the horizontal. hence, 90-60 is equal to 30degress. do you think that makes sense...yea..tis kinda hard to explain unless i draw you a figure....but plz check if my ans. is right.....

5. Jul 3, 2006

### vijay123

from then on, i can show you my working

6. Jul 3, 2006

### jhrnndz1

I'm sorry, but I don't have the answer key to check your answer.

7. Jul 3, 2006

### Hootenanny

Staff Emeritus
I suggest you start this question by drawing a diagram. You can use trig to find both the angle of launch and the initial height above the ground. From here you should be able to solve for the flight time an hence the displacement from the base. Remember that, (a) there is no acceleration in the horizontal plane; (b) don't forget to add on the additional 1.75m once you have found the initial height of the keys above the bottom of the wheel.

8. Jul 3, 2006

### vijay123

oh den, but do you knwo how i got the angle?

9. Jul 3, 2006

### vijay123

once you know how the angle wus got, then as how hootnanny says, you can use trignomertry to solve the question.....yea and it would be very useful drawing a diagram of a point at 10 o clock(just like your clok)and join it to the circle origin.

10. Jul 3, 2006

### vijay123

then, if 6ocloclk is 180degress, and 10 o clock is 300degress.360-300=60degrss...from then on, i think you can handle it.

11. Jul 3, 2006

### jhrnndz1

okay, so is it correct that the total height is 5m from the radius at the bottom of the ferris wheel, then 5msin30 + the base (1.75) = 9.25

12. Jul 3, 2006

### jhrnndz1

Then i used the y=(vosintheta)t-1/2gt^2 to solve for time, which i got 1.4seconds, and then put that in the equation x=(v0costheta)t and calculated x=1.18meters

13. Jul 3, 2006

### vijay123

no, you need to add 5m more the the heght, coz you need to add in the radius of the circle....i mean 5sin30+5+1.75=11.0801.

14. Jul 3, 2006

### vijay123

nooooooooooo.....worong...the angle is 60 degress not 30!!!!!!!!!!

15. Jul 3, 2006

### vijay123

5sin60+5+1.75=11.0801

16. Jul 3, 2006

### jhrnndz1

Was it correct the way i solved the question, except for the angle mistake?

17. Jul 3, 2006

### Hootenanny

Staff Emeritus
I hesitate to intrude here, but just to clarify some things. The 5sin60 gives you the vertical height above the centre of your wheel. To obtain the height above the ground you must first add the distance from the centre of the wheel the the bottom of the wheel (the radius) and then the distance from the base of the wheel to the floor.

18. Jul 3, 2006

### Hootenanny

Staff Emeritus
Your approach was indeed correct, but take note of the displacement corrections mentioned above.

19. Jul 3, 2006

### vijay123

yes....see...you add radius of circle+height beneath+the remaining segment.

20. Jul 3, 2006

### vijay123

this is beacuse.....the 60 degress from the triangle....the triangle is formed when you join the orgin to the point at 10 o clock. then you draw a line vertically downwardsstaright from the point and form another line horizontal to the radius...
wallah you get a trianle...