What is the Correct Direction of Velocity in the Texas Tumblers Circus Act?

In summary, the conversation discussed a physics problem involving a circus act. The initial velocity of the performer, Mary Belle, was calculated at 14 m/s at an angle of 53 degrees. The problem required finding the magnitude and direction of Mary Belle's velocity when she reaches Joe Bob's hands, which are 6.1 m above and 8.2 m horizontally from her launch point. The correct answers were 8.4 m/s for the magnitude and 9.1 degrees above the horizontal for the direction. The conversation also mentioned the difficulties with the online system, Mastering Physics, and the importance of not rounding off numbers.
  • #1
mnafetsc
14
0

Homework 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.

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

Homework Equations



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

The Attempt at a Solution



arctan=(v_y/v_x) which gives me 7.9 degrees which is wrong
 
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  • #2
mnafetsc said:
Additionally I have v_0 is 14 m/s, v=8.4 m/s, v_x= 8.42, v_y=1.17
How did you calculate the velocity? How can v_x, a component, be greater than v?
 
  • #3
sqrt)v_x^2+v_y^2

my actual answer was 8.5 but mastering physics rounded it down to 8.4
 
  • #4
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
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
mnafetsc said:
v_x=14cos53= 8.42
This looks OK. (Assuming I understand the problem.)

V_y=13sin53- 9.8(8.6/8.42)=1.17
This seems to have a few errors. (At least going by the info in your first post.)

The data seems a bit off.
 
  • #7
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
mnafetsc said:
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
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
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?

correct answer is 14 m/s

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

correct answer 8.4 m/s

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

part I am 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 I am doing on masteringphysics
 
  • #10
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
Doc Al said:
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.

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.
 

1. What is velocity and how is it different from speed?

Velocity is a measure of an object's displacement over time, taking into account both the magnitude and direction of movement. Speed, on the other hand, only measures the magnitude of an object's movement. In other words, velocity tells us how fast and in what direction an object is moving, while speed only tells us how fast it is moving regardless of direction.

2. How is the direction of velocity represented?

The direction of velocity is typically represented by an arrow pointing in the direction of movement. The length of the arrow represents the magnitude of the velocity, while the direction it is pointing in represents the direction of movement. Alternatively, the direction of velocity can also be represented using positive and negative signs, with positive indicating movement in one direction and negative indicating movement in the opposite direction.

3. Can an object have a constant velocity?

Yes, an object can have a constant velocity if it is moving at a constant speed in a straight line. This means that the object's direction and speed do not change over time, and the velocity remains the same. However, any change in speed or direction results in a change in velocity, even if the magnitude of the change is very small.

4. How is the direction of velocity affected by forces?

The direction of velocity is affected by forces acting on an object. If a force acts in the same direction as the velocity, it will cause the object to speed up in that direction. If the force acts in the opposite direction, it will cause the object to slow down. If the force acts perpendicular to the velocity, it will cause the object to change direction but not its speed. In all cases, the direction of the velocity is influenced by the direction of the force acting on the object.

5. How do you calculate the direction of velocity?

The direction of velocity can be calculated using trigonometry. For example, if an object is moving at a velocity of 20 m/s at an angle of 30 degrees from the horizontal, we can use the formula tangent (θ) = opposite/adjacent to calculate the direction of velocity. In this case, the direction of velocity would be 30 degrees from the horizontal in the direction of the object's movement.

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