How Is Force Calculated in a Human Cannonball Stunt?

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Homework Help Overview

The problem involves calculating the force exerted on a performer in a human cannonball stunt, where the performer is launched from a cannon at an angle and travels a specified distance. The context includes parameters such as mass, distance, launch angle, and acceleration, while neglecting air resistance.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Newton's laws and projectile motion equations to determine the force. Some express confusion over the relationship between acceleration and velocity, while others suggest focusing on the trajectory and kinetic energy.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to relate force, acceleration, and launch speed. Some have provided equations and reference frames, while others have noted the need for clarity in assumptions and calculations. There is no explicit consensus on the method to be used.

Contextual Notes

Participants mention constraints such as the requirement to avoid energy transformations and the need for clarity in the reference frames used for calculations. There is also a note about the unreadability of shared images, which may hinder the discussion.

Arman777
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Homework Statement


The Zacchini family was renowned for their human-cannonball act in which a family member was shot from a cannon using either elastic bands or compressed air.In one version of the act,Emanuel Zacchini was shot over three Ferris wheels to land in a net at same height as the open end of the cannon and at a range of 69 m.He was propelled inside the barrel for 5.2 m and launched at an angle of 530 ( degree).If his mass was 85 kg and he underwent cosntant acceleration inside the barrel,what was the magnitude of the force propelling him ? ( Hint:Treat the launch as though it were along aa ramp at 530 ( degree).Neglect air drag )

Homework Equations


\vec F_t=m \vec a m=mass \vec F_t =Total \ Force
\vec w_=m \vec g g=9.8 m/s^2
v-v_0=at v_0=inital \ velocity

Projectile motion equations (In this question they will be in different forms I guess and I don't know how to find them that's my problem)



The Attempt at a Solution


[/B]
I upload a picture.After that solution I am stucked.Cause there's acceleration and also velocity of the object and I have to apply them to projectile motion equations.

Here my try
For x-axis ;

69m=v_0cos(53^0) t+\frac 1 2 acos(53^0)t^2
Simply I couldn't solve.

For y axis;
Here things are more complicated
Simply I couldn't solve.

Also another referance frame would be easier to solve but I couldn't put the forces in there something's doesn't make sense.I would be happy If somebody help me
Thanks
 

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Just FYI, your image is ridiculously small and completely unreadable.
 
phinds said:
Just FYI, your image is ridiculously small and completely unreadable.
Then this is the best thing I can do

Here F is The force that cannonball applies, I showed my referance frame, m is mass of object, a is the acceleration of object inside the cannonball.
 

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In this referance frame I can write the components of force.But is it a better then my pre-referance frame...
Components are ;
w=mg (-y)
N=Ncos37 (x) + Nsin37(y)
F=Fcos53 (-x) + Fsin53(y)

so Ftotal=ma cos53 (-x) + ma sin53 (y)

so ;
-ma cos53=Ncos37-Fcos53
and
ma sin53 =-mg+Nsin37+Fsin53
 

Attachments

  • Adsız 11.png
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Deal with the trajectory first. Find the KE with which he must emerge from the barrel.
 
I can use only Newtons's Law.
And I just don't want to solve the question and leave there.I want to know where am I doing wrong.
Thanks
 
Arman777 said:
I can use only Newtons's Law.
And I just don't want to solve the question and leave there.I want to know where am I doing wrong.
Thanks
You will have to deal with the trajectory at some point, and that also uses Newton's laws.
But if you wish to start with the cannon, ok.
It will be much easier if you use a coordinate parallel to the ramp. Try writing the F=ma equation for that.
You won't need to work with a second coordinate.
 
I didnt want to use energy cause the book hasnt arrived there yet.So this question has a solution without using any energy transformation.(If that was your point)
So there's two equations.(there's a pic in my #3 post A simple question about Force )
##-ma = - F + mgsin(53^0) ## (Eq-1)
##F:\text{The force that we wanted to know}##
##a:\text{acceleration of object}##
##N=mgcos(53^0)## (Eq-2)
 
Arman777 said:
I didnt want to use energy cause the book hasnt arrived there yet.So this question has a solution without using any energy transformation.(If that was your point)
So there's two equations.(there's a pic in my #3 post A simple question about Force )
##-ma = - F + mgsin(53^0) ## (Eq-1)
##F:\text{The force that we wanted to know}##
##a:\text{acceleration of object}##
##N=mgcos(53^0)## (Eq-2)
Ok, that relates F to a. Next, you need to relate a to the launch speed. What equations do you know that would do that?
 
  • #10
##v^2-(v_0)^2=2a(x-x_0)## So here ##v_0=0## and ##x_0=0##
so If we use this equation to in our referance frame we willl get

##v^2-0=2(-a)((-x)-0)## which equals to;
## v^2=2ax##
 
  • #11
Ok I solved the problem.Just I made a simple false assumption.And then I couldn't solve cause of that.Now I solved thanks again
 

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