Acrobats trajectory with a swing

In summary, the problem involves two acrobats launching themselves from a swing with a velocity of 6.80 m/s and an angle of 32.0 degrees upwards. At the top of their trajectory, one acrobat pushes the other to reach another swing at the same height. To find the distance between the two swings, the maximum height reached by the acrobats must be calculated, and the force needed to make one acrobat fall straight downwards must be determined. This force can then be used to calculate the horizontal distance traveled by the other acrobat, which, when added to the distance from the first swing to the maximum height, gives the total distance between the two swings.
  • #1
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Homework Statement



Two acrobats, each of 72.0 kg, launch themselves together from a swing, holding hands. Their velocity at launch was 6.80 m/s and the angle of their initial velocity relative to the horizontal was 32.0 degrees upwards. At the top of their trajectory, their act calls for them to push against each in such a way that one of them becomes stationary in mid-air and then falls to the safety net while the other speeds away, to reach another swing at the same height at that of the swing they left. What should be the distance (in m) between the two swings?

Homework Equations





The Attempt at a Solution



I can calculate the maximum height they reached, but what I dint understand is when they push each other ...

thanks for the help
 
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  • #2
The question does seem a little bit vague but I think I understand. At the maximum height there is only velocity in the horizontal direction. So when acrobat A pushes off of acrobat B, acrobat B loses all of his/her horizontal velocity and falls straight downwards while acrobat A gets a nice boost to the next swing. You will basically need to find out the force that acrobat A needs to apply to acrobat B to make acrobat B fall straight downwards. Then use that same force to find out how much farther acrobat B will go.
 
  • #3
if you haven't found the answer yet, steps to follow:

separate V into Vx and Vy

for the vertical component
use (Vy2-Vy1)/a=t to figure out t
this is the time it takes to get from swing 1 to max height

for horizontal component
d=Vxt find d
(note the t is the same as the t you calculated above)
this is the horizontal distance from swing one to when max height occurs

now use momentum equation. specifically:

Vx(m1+m2)=m1v1
since m1=m2

Vx(2m)=mv1
v1=2Vx

use the horizontal component equation of

D=Vxt to find D (again).
this time, Vx=v1 (horizontal velocity has changed) so d=v1t

but if you realize this time, the D is horizontal distance from when max height occurs to the second swing.

to answer your question, add d and D.

tada.

good luck with your CAPA
 

What is an acrobat's trajectory with a swing?

An acrobat's trajectory with a swing refers to the path or trajectory that an acrobat follows when performing a swing on a trapeze or other apparatus.

How does an acrobat's trajectory with a swing differ from other types of movements?

An acrobat's trajectory with a swing is unique in that it involves a combination of swinging and flying movements, as well as precise timing and coordination.

What factors affect an acrobat's trajectory with a swing?

The height and speed of the swing, the strength and technique of the acrobat, and the weight and design of the apparatus can all affect an acrobat's trajectory with a swing.

Can an acrobat control their trajectory with a swing?

Yes, experienced acrobats are able to control their trajectory with a swing by adjusting their body position, timing their movements, and using their strength and technique to navigate the swing.

What are the risks associated with an acrobat's trajectory with a swing?

As with any acrobatic movement, there is always a risk of falling or injury when performing a trajectory with a swing. It is important for acrobats to have proper training and safety measures in place to minimize these risks.

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