- #26

ehild

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I attached a figure to my previous post, look at it.

What is the equation between the x and y coordinates of the projectile?

ehild

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- #26

ehild

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I attached a figure to my previous post, look at it.

What is the equation between the x and y coordinates of the projectile?

ehild

- #27

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what is the equation? what formula are you using to get the initial velocity?

- #28

ehild

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That you have to know. What equations are there for projectile motion?

ehild

ehild

- #29

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- #30

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Vo = V - at

Vo - initial velocity

V - Velocity

a - acceleration

t - time

- #31

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1. Given range(distance first and last wheel). From this you can calculate horizontal the velocity V

2. From this velocity you find the maximum height the man goes above the middle wheel assuming equally spaced wheels.

3. Add this height to the height from top of the wheel to the gun.

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

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- #33

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where are you masters???

- #34

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- #35

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can anyone solve this?

- #36

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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

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is there an equation involve here?

- #38

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You certainly learnt 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

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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

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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

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why are we talking abouit energy now? im getting more confuse... oh my god... help!!!

- #42

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- #43

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voko thank you... but i really dont understand hehehe sorry guys...

- #44

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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

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can anyone give me the formula's to use?

- #46

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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}

V^{2} xtan[θ] Cos[θ]^{2} =(1/2)g x^{2}

V^{2} tan[θ] Cos[θ]^{2} =(1/2)g x

For optimum range, θ=45°

Range x=V^{2}/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

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])

xtan[θ]=(1/2)g (x/(VCos[θ])

V

V

For optimum range, θ=45°

Range x=V

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)

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- #47

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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

- #48

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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.

- #49

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This google search will also help:

http://www.google.com/search?q=find+a+parabola+using+three+points

http://www.google.com/search?q=find+a+parabola+using+three+points

- #50

ehild

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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 V_{x} and y component V_{y}. 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 Y_{max}= V_{y}^{2}/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/V_{x}. Its height changes from Y_{max} to H, and the vertical velocity component becomes gΔt. Write up conservation of energy: You get an equation for V_{y} in terms of V_{x}. To ensure minimum Vo, the derivative of V_{x}^{2}+V_{y}^{2} 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

The human cannon ball is launched at Vo velocity, with x component V

The ball moves along a parabola, and the highest point above the middle wheel is Y

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/V

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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