How to get initial and final velocity? need help thanks

  • #26
ehild
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You have to find out both the initial velocity and the distance of launch place from the first wheel.
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
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
 
  • #29
but i still have to get the time for me to calculate initial velocity right? then how can i get the time? why is this sooo confusing... sorry guys...
 
  • #30
is this the right formula for getting the initial velocity?
Vo = V - at
Vo - initial velocity
V - Velocity
a - acceleration
t - time
 
  • #31
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but i still have to get the time for me to calculate initial velocity right? then how can i get the time? why is this sooo confusing... sorry guys...

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

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
 
  • #32
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What we have been saying is that you cannot calculate anything now, because the problem requires that you make certain assumptions on the arrangement. All you know is how wide the three wheels (= W) to fly over, and how high (= H) they are. But you do not know how far from the wheels the launch pad and the target net are. These unknown distances and W and H will together result in certain restrictions on the equations of motion, from which you could deduce the minimum initial velocity.
 
  • #33
where are you masters???
 
  • #34
so your saying il make assumptions so theres no wrong answer here? outcome will depend on my assumptions?
 
  • #35
can anyone solve this?
 
  • #36
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so your saying il make assumptions so theres 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
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
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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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
why are we talking abouit energy now? im getting more confuse... oh my god... help!!!
 
  • #42
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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
voko thank you... but i really dont understand hehehe sorry guys...
 
  • #44
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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
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
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
 
Last edited:
  • #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 not to shoot as far as you can while overflying an obstacle.
 
  • #48
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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
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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 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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