How to get initial and final velocity? thanks

[jamesblim168]

Homework Statement

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. Assuming that he missed the net and landed on the ground, with what velocity will he strike the ground?

Homework Equations

The Attempt at a Solution

 

[jamesblim168]

up for today...

 

[jamesblim168]

up for today...

 

[voko]

If no more conditions are given, it seems that the problem requires that you find an optimal setup for the stunt, optimal in the sense that the initial velocity is minimal.

 

[jamesblim168]

If no more conditions are given, it seems that the problem requires that you find an optimal setup for the stunt, optimal in the sense that the initial velocity is minimal.

i don't understand how... can you teach me how?

 

[HallsofIvy]

You have made no attempt at all, not even citing equations that might be relevant. Are you saying that you are taking a course in which this was given as an exercise and have learned nothing about this?

 

[voko]

i don't understand how... can you teach me how?

Imagine it is you who will be fired from a cannon in order to fly over three ferris wheels. You are in full control of the arrangement. How would you approach the problem?

 

[jamesblim168]

1. Homework Statement
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. Assuming that he missed the net and landed on the ground, with what velocity will he strike the ground?


2. Homework Equations
my problem i don't know what equation il use... coz there so many missing values
is the initial velocity is 0?3. The Attempt at a Solution
this is what i did, i know its not right i hjope you guys could help me out...
“Y=Viy * t + 0.5 * Ay * t^2 “ to solve for time
-20.50m = (0 m/s) * t + 0.5 (-9.8 m/s/s) * t^2
-20.50m = (-4.9 m/s/s) * t^2
4.1s^2 = t^2
(TIME) T= 2.02s

Answer:
Problem 1:Initial velocity is 0

Problem 2: Assuming that he missed the net and landed on the ground, with what velocity will he strike the ground?

V= distance/time
V=10.63/2.02s
V=5.26m/s

 

[jamesblim168]

up for today...

 

[jamesblim168]

up for today...

 

[azizlwl]

It depends on the horizontal distance of the cannon and the first wheel, and the last wheel with the safety net.

 

[jamesblim168]

how would i know? those are the only values i have... does this mean this problem cannot be solve?

 

[Jakeus314]

An initial velocity of zero makes no sense. He wouldn't go anywhere. Can you decide where he might have to be at some point in space? Like as he approaches the first wheel?

 

[Jakeus314]

how would i know? those are the only values i have... does this mean this problem cannot be solve?

You may need to make an assumption about the placement of the first Ferris wheel and its distance from the cannon.

 

[jamesblim168]

what do you mean? how to decide?

 

[jamesblim168]

can anyone help me solve this?

 

[jamesblim168]

would really appreciate the effort ... thanks

 

[Jakeus314]

Is a drawing included? The orientation and position of the wheels will be important. If no drawing is included you will need to start the problem making an assumption about where the wheels are.

 

[azizlwl]

You can find the velocity for maximum range, the distance between the wheels.
For maximum range the angle at above first wheel is 45°.

 

[jamesblim168]

up for today...

 

[Jakeus314]

What is "up for today.." ?

 

[azizlwl]

up for today...

Can you find the velocity just above the first wheel?

 

[jamesblim168]

Can you find the velocity just above the first wheel?

no, coz i don't know how...

 

[ehild]

Read about "projectile motion". The man moves both in horizontal and vertical direction. There is an equation for the x and y coordinates of the trajectory.
The projectile has to fly over the three ferris wheels, 18 m high and with distance 10.63 m between the first and last ones. The man had to be at least 18 m high above the ground at the place both of the first and last wheels.
Make a drawing to show the launch place, the wheels and the trajectory of the human cannonball.
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.

ehild

 

[jamesblim168]

so what is my initial veloctiy if my angle is 45 degress?

 

[ehild]

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

 

[jamesblim168]

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

 

[ehild]

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

ehild

 

[jamesblim168]

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...

 

[jamesblim168]

is this the right formula for getting the initial velocity?
Vo = V - at
Vo - initial velocity
V - Velocity
a - acceleration
t - time

 

[azizlwl]

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 V_x.

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

 

[voko]

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.

 

[jamesblim168]

where are you masters?

 

[jamesblim168]

so your saying il make assumptions so there's no wrong answer here? outcome will depend on my assumptions?

 

[jamesblim168]

can anyone solve this?

 

[voko]

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.

 

[jamesblim168]

is there an equation involve here?

 

[voko]

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.

 

[jamesblim168]

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

 

[voko]

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?

 

[jamesblim168]

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

 

[voko]

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.

 

[jamesblim168]

voko thank you... but i really don't understand hehehe sorry guys...

 

[voko]

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.

 

[jamesblim168]

can anyone give me the formula's to use?

 

[azizlwl]

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

 

[voko]

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.

 

[Jakeus314]

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.

 

[Jakeus314]

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

 

[ehild]

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_y²/2g.

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_x²+V_y² 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