How Is Force Calculated in a Human Cannonball Stunt?

[Arman777]

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

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

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

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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 ;

[tex]69m=v_0cos(53^0) t+\frac 1 2 acos(53^0)t^2 [/tex]
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

 

[phinds]

Just FYI, your image is ridiculously small and completely unreadable.

 

[Arman777]

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.

 

[Arman777]

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 F_total=ma cos53 (-x) + ma sin53 (y)

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

 

[haruspex]

Deal with the trajectory first. Find the KE with which he must emerge from the barrel.

 

[Arman777]

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

 

[haruspex]

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.

 

[Arman777]

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.(Theres 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)

 

[haruspex]

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.(Theres 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?

 

[Arman777]

##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##

 

[Arman777]

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