# Direction of velocity

1. Jun 4, 2010

### mnafetsc

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

A new circus act is called the Texas Tumblers. Lovely Mary Belle swings from a trapeze, projects herself at an angle of 53 ^\circ, and is supposed to be caught by Joe Bob, whose hands are 6.1 {\rm m} above and 8.2 {\rm m} horizontally from her launch point (the figure ). You can ignore air resistance.

For the initial speed calculated in part A, what is the direction of her velocity when Mary Belle reaches Joe Bob?

Additionally I have v_0 is 14 m/s, v=8.4 m/s, v_x= 8.42, v_y=1.17

2. Relevant equations

Im using arctan=(v_y/v_x) which gives me 7.9 degrees which is wrong

3. The attempt at a solution

arctan=(v_y/v_x) which gives me 7.9 degrees which is wrong

2. Jun 4, 2010

### Staff: Mentor

How did you calculate the velocity? How can v_x, a component, be greater than v?

3. Jun 4, 2010

### mnafetsc

sqrt)v_x^2+v_y^2

my actual answer was 8.5 but mastering physics rounded it down to 8.4

4. Jun 4, 2010

### Staff: Mentor

I meant how did you solve for v based on the initial velocity? How did you solve for v_y? (I get different values for v and v_y, and thus for the angle.)

5. Jun 4, 2010

### mnafetsc

v_x=14cos53= 8.42

V_y=13sin53- 9.8(8.6/8.42)=1.17

These two answers then gave me 8.4 according to mastering physics, which was the answer to part B of this problem

I got t for the v_y equation from x=v_0 cos (theta) t
which I used to help me find the initial velocity for part A of this problem.

6. Jun 4, 2010

### Staff: Mentor

This looks OK. (Assuming I understand the problem.)

This seems to have a few errors. (At least going by the info in your first post.)

The data seems a bit off.

7. Jun 4, 2010

### mnafetsc

I meant 14 but I did replace that 8.6( bad reading on my part) with 8.2 which gave me a new v_y of 1.64

so arctan=(1.64/8.42) gives me 11 degrees above the horizontal which is still wrong

8. Jun 4, 2010

### Staff: Mentor

To delve into things further, I'll need to know the complete statement of the problem. If I understand you, the initial velocity is 14 m/s at an angle of 53 degrees with the horizontal. From that point on it's a projectile that's supposed to reach +6.1 for y and + 8.2 for x. Why don't you check and see if that's possible? You figured out the the time based on the horizontal distance traveled. What height would it reach in that time?

(Is this a problem from a textbook? Or just on the online system?)

9. Jun 4, 2010

### mnafetsc

This is the complete problem statement:

A new circus act is called the Texas Tumblers. Lovely Mary Belle swings from a trapeze, projects herself at an angle of 53 ^\circ, and is supposed to be caught by Joe Bob, whose hands are 6.1 {\rm m} above and 8.2 {\rm m} horizontally from her launch point (the figure ). You can ignore air resistance.

part A:What initial speed v_0 must Mary Belle have just to reach Joe Bob?

part B:For the initial speed calculated in part A, what is the magnitude of her velocity when Mary Belle reaches Joe Bob?

part C:For the initial speed calculated in part A, what is the direction of her velocity when Mary Belle reaches Joe Bob?

part Im stuck at

Also, I plugged in time I got and it gave me y= 6.11 so roughly the same as the given
This is a book problem Im doing on masteringphysics

10. Jun 4, 2010

### Staff: Mentor

OK, I found this problem in Young and Freeman. What you'll need to do is not round anything off (except when giving your answers). So recalculate the initial velocity to at least 3 significant figures (don't round off to 14) for the purpose of doing the other parts of the problem. Then you'll probably do OK. Slight inaccuracies make a big difference in the angle.

You're not the first to have a problem with Mastering Physics. It can be very picky.

11. Jun 4, 2010

### mnafetsc

Haha just to be safe I decided not to round at all and got the answe 8.9 still wrong, but I used 3 sigfigs and got 9.2 which it accepted for 9.1 thanks for your help.

Im just glad to know it was mastering physics this time and not me.