How to get initial and final velocity? thanks

  • #36
voko
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391
so your saying il make assumptions so there's no wrong answer here? outcome will depend on my assumptions?

If you make some assumptions and get an answer, that will be a very good start. To finish off, you then just need to analyse how your assumptions affect the initial velocity, and change the assumptions to make it minimal.

For example, you could assume that the launch pad is ten miles away from the wheel. And you will get an answer. But the initial and terminal velocities will be so high that you might as well call the human cannonball a meatball.
 
  • #37
jamesblim168
31
0
is there an equation involve here?
 
  • #38
voko
6,054
391
You certainly learned that you get the maximum distance if you launch a projectile at 45 degrees. The human projectile certainly used that launch angle.

4. You have total height to cover with 45degree projectile

The 45 degree elevation is not optimal here. The objective is not to have the maximum distance covered, but have minimum energy with some restrictions on the shape of the trajectory. The optimal angle depends on the ratio of the wheel dimensions.
 
  • #39
jamesblim168
31
0
The 45 degree elevation is not optimal here. The objective is not to have the maximum distance covered, but have minimum energy with some restrictions on the shape of the trajectory. The optimal angle depends on the ratio of the wheel dimensions.

now its just got more complicated... hehehe can you just solve it then explain it to me why? really appreciate it voko... thanks
 
  • #40
voko
6,054
391
OK, I will provide a couple of hints.

1. The arrangement must be symmetrical, i.e., the distance from the launch pad to the first wheel and from third wheel to the target net must be equal. If that is not the case, the trajectory will not be optimal for energy. You need to think why this is so.

2. Let's say the cannonball is at the apex of the trajectory right over the middle of the second wheel. What can be said about its total energy?
 
  • #41
jamesblim168
31
0
why are we talking abouit energy now? I am getting more confuse... oh my god... help!
 
  • #42
voko
6,054
391
We are talking about the energy because it is the energy that converts a human cannonball into a meatball. We want the least energy possible.
 
  • #43
jamesblim168
31
0
voko thank you... but i really don't understand hehehe sorry guys...
 
  • #44
voko
6,054
391
The total energy is constant during the entire flight.

At the apex, it has a particularly simple form that you can use to meet the constraints of the task (not flying into wheels) and obtain the minimal energy. From that, everything else can be determined very easily.
 
  • #45
jamesblim168
31
0
can anyone give me the formula's to use?
 
  • #46
azizlwl
1,065
10
Let x be the distance from first wheel to the last wheel.
x=VCos[θ]t
t=x/(VCos[θ]

Let y be the vertical displacement above the wheel.
y=VSin[θ]t-(1/2)g(x/(VCos[θ])2
xtan[θ]=(1/2)g (x/(VCos[θ])2
V2 xtan[θ] Cos[θ]2 =(1/2)g x2
V2 tan[θ] Cos[θ]2 =(1/2)g x

For optimum range, θ=45°
Range x=V2/g

Horizontal velocity Vx=VCos[45°], it remains constant for the whole flight.
Vertical velocity Vy=VCos[45°], this changes in flight.
From above we have the value of vertical velocity, Vyf, just above the first wheel.

We can get initial velocity where we know vertical distance travelled, from the gun's level to the top first wheel's level.

|Vyf|=|Vx|
(Vyf)2=(Vyi)2-2as
 
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  • #47
voko
6,054
391
For optimum range, θ=45°

Again, θ=45° is not for optimum range. It's for maximum range. Which has nothing to do with the problem at hand. The problem is not to shoot as far as you can while overflying an obstacle.
 
  • #48
Jakeus314
47
0
can anyone give me the formula's to use?

"the formulas" are the projectile motion equations.
Learn what they all mean...

There is no single formula for this problem, at least not what I'm guessing you want. (students ask all the time for one.. for problems like this)
http://en.wikipedia.org/wiki/Projectile_motion

You can do it. I hope for your sake you buckle down and really visit that link.

And ignore the 45 degree stuff. That is a specific problem type that is related to but is not necessary for this problem.

I also still think that info is missing from the problem description. It still can be solved, given that you make some assumptions and move on.
 
  • #50
ehild
Homework Helper
15,543
1,915
It is reasonable to assume that the wheels are adjacent, as shown in the figure. D=10.63 m, the launch happens at 2.50 m height, so the apex of the 18 m height wheels are H=15.5 m above the launch position.

The human cannon ball is launched at Vo velocity, with x component Vx and y component Vy. You have to find the minimal Vo which ensures that the ball is at height H above the first and last wheels.

The ball moves along a parabola, and the highest point above the middle wheel is Ymax= Vy2/2g.

Consider the time instant when the ball is at the apex of the parabola, above the middle wheel. The horizontal velocity component is constant during the flight. It reaches the last wheel in time Δt=D/Vx. Its height changes from Ymax to H, and the vertical velocity component becomes gΔt. Write up conservation of energy: You get an equation for Vy in terms of Vx. To ensure minimum Vo, the derivative of Vx2+Vy2 has to be zero.

The text of the problem can be understood that the human cannon ball has to fly over the distance spanned by the three wheels. In this case, the ball was fired at the edge of the first wheel, D/2 distance from the apex, at height of 2.5 m. Using the equation of the trajectory, Vx and Vy has to be found which correspond y=15.5 m at both x1=D/2 and x2=5D/2.

ehild
 

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  • #51
CAF123
Gold Member
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The way I thought of doing it initially was to use the well known trajectory equation [tex] y = x tanθ - \frac {g x^2}{2 v_o^2 cos^2 θ} [/tex] and put in values for [itex] y [/itex] and [itex] x [/itex] depending on where the human cannonball is at specific stages of the journey.
For example, we want, at say [itex] x = 3m, y [/itex] to be 15.50m. Here, my assumption is that 3m is the distance from the first wheel to the cannon. The only unknowns you have is [itex] v_o [/itex] and [itex] θ [/itex].

I then took another stage in the journey say, right at the end, where [itex] y = 2.50 m [/itex] and [itex] x [/itex] is about [itex] (10.63 +6) ≈ 17m [/itex]. Thus, I have two eqns and two unknowns which can be solved to yield some [itex] θ [/itex] and [itex] v_o [/itex].

Is this method plausible? Of course, after getting some result, to get minimal velocity, you could tweak the distance from the cannon to the first wheel and see what happens.
 
  • #52
ehild
Homework Helper
15,543
1,915
Yes, you can try and see when you get the minimum vo. But the problem has an exact solution.
The geometry of the problem is not clearly stated. One can understand that the three wheels altogether cover 10.63 m distance, or each of them does.

a human cannonball in 1940 soared over three ferris wheels, each 18 meters high covering a horizontal distance of 10.63 meters. Assuming that the point of projection is 2.50 meters above the ground and that he landed safely on a net placed at the same level, find his initial velocity.

Looking at the picture of a Ferris Wheel, I am inclined to believe the second geometry is valid.

ehild
 

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  • #53
voko
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391
Is this method plausible? Of course, after getting some result, to get minimal velocity, you could tweak the distance from the cannon to the first wheel and see what happens.

Technically, it should work. In practice, it seems to result in very messy algebra. The method of looking at the energy state at the apex, as detailed by ehild, gets you to the result much quicker.
 
  • #54
CAF123
Gold Member
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I would agree, but it is nice to know there are other ways to tackle things.
 

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