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Tennis Ball Launcher Calcuations

by Redfire66
Tags: calcuations, tennis ball launcher
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Redfire66
#1
Jan7-13, 03:09 PM
P: 5
1. The problem statement, all variables and given/known data
I'm building a tennis ball launcher (as most people did) and I barely have some calculations done. I already have a measurement for my project (i.e. an angle and such), a k value for my string and a distance that it should travel horizontally. Theoretically, I'm supposed to calculate the velocity it takes the ball to travel the distance horizontally and a stretch distance (x) for my string; I'm not given a maximum height (but it should be above 1m), or time but I do have an initial height where the ball is launched (which is the height of my launcher) - there's quite a bit of missing variables without actually simulating it (since it's required that I calculate before I simulate)

2. Relevant equations
Dx=vtcosθ
Fg=kx
E=E
Dy=vtsinθ
V=vcosθ or sinθ depending on direction
(I might be missing some but I suppose you'd get the idea)

3. The attempt at a solution
Determined the k value by determining the force (mg) and relating to the F=kx equation
I've tried to do Et=Et
But I end up getting stuck with
E(at launch) = E (when pulled back)
mgh + 1/2mv^2 = 1/2kx^2 + mgh
since the heights are different, and since I don't know how far back I stretch it (x), then I won't be able to determine how high it is (for the mgh on the second half); maybe I did it wrong or used the wrong method, but that's how I seem to think I should do it
I'm still thinking it through how I'm supposed to attempt this, and seems sort of difficult without being given more variables
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CWatters
#2
Jan7-13, 05:00 PM
P: 3,135
Theoretically, I'm supposed to calculate the velocity it takes the ball to travel the distance horizontally
You don't need energy equations for that. I would use equations of motion. Start by working out the flight time using the vertical velocity and the equations of motion.
berkeman
#3
Jan7-13, 05:12 PM
Mentor
berkeman's Avatar
P: 40,926
Quote Quote by Redfire66 View Post
1. The problem statement, all variables and given/known data
I'm building a tennis ball launcher (as most people did) and I barely have some calculations done. I already have a measurement for my project (i.e. an angle and such), a k value for my string and a distance that it should travel horizontally. Theoretically, I'm supposed to calculate the velocity it takes the ball to travel the distance horizontally and a stretch distance (x) for my string; I'm not given a maximum height (but it should be above 1m), or time but I do have an initial height where the ball is launched (which is the height of my launcher) - there's quite a bit of missing variables without actually simulating it (since it's required that I calculate before I simulate)

2. Relevant equations
Dx=vtcosθ
Fg=kx
E=E
Dy=vtsinθ
V=vcosθ or sinθ depending on direction
(I might be missing some but I suppose you'd get the idea)

3. The attempt at a solution
Determined the k value by determining the force (mg) and relating to the F=kx equation
I've tried to do Et=Et
But I end up getting stuck with
E(at launch) = E (when pulled back)
mgh + 1/2mv^2 = 1/2kx^2 + mgh
since the heights are different, and since I don't know how far back I stretch it (x), then I won't be able to determine how high it is (for the mgh on the second half); maybe I did it wrong or used the wrong method, but that's how I seem to think I should do it
I'm still thinking it through how I'm supposed to attempt this, and seems sort of difficult without being given more variables
Welcome to the PF.

If you check out the bottom of this page, you will see some other "tennis ball launcher" threads that may also help you out.

BTW, air resistance is not negligible for a tennis ball. Have you been told to ignore it anyway for this project? If not, there are some reasonable info pages on factoring air resistance into projectile calcs at wikipedia.org

Redfire66
#4
Jan7-13, 05:33 PM
P: 5
Tennis Ball Launcher Calcuations

Quote Quote by CWatters View Post
You don't need energy equations for that. I would use equations of motion. Start by working out the flight time using the vertical velocity and the equations of motion.
Well other than the fact that I need to calculate velocity, I'm not sure of any other method to calculate the stretch distance for string (x). Maybe I forgot an equation?
Quote Quote by berkeman View Post
Welcome to the PF.

If you check out the bottom of this page, you will see some other "tennis ball launcher" threads that may also help you out.

BTW, air resistance is not negligible for a tennis ball. Have you been told to ignore it anyway for this project? If not, there are some reasonable info pages on factoring air resistance into projectile calcs at wikipedia.org
Yes, I am neglecting air resistance. Sorry for not mentioning; am most of the other projects that I am aware of don't have some of the information I'm looking for.
CWatters
#5
Jan7-13, 06:12 PM
P: 3,135
Quote Quote by Redfire66 View Post
Well other than the fact that I need to calculate velocity, I'm not sure of any other method to calculate the stretch distance for string (x). Maybe I forgot an equation?
One step at a time. Have you already calculated the launch velocity required?

Then regarding....

E(at launch) = E (when pulled back)
mgh + 1/2mv^2 = 1/2kx^2 + mgh
I would redefine the origin as the height when pulled back so it becomes..

E(at launch) = E (when pulled back)
mgh + 1/2mv^2 = 1/2kx^2 + 0

Then presumably the elastic is pulled back at the launch angle so

h = xsinθ
Redfire66
#6
Jan7-13, 09:20 PM
P: 5
Quote Quote by CWatters View Post
One step at a time. Have you already calculated the launch velocity required?

Then regarding....



I would redefine the origin as the height when pulled back so it becomes..

E(at launch) = E (when pulled back)
mgh + 1/2mv^2 = 1/2kx^2 + 0

Then presumably the elastic is pulled back at the launch angle so

h = xsinθ
Actually no, I have no calculated the initial velocity, which is why I sort of went to check how I would approach the others. I guess it just adds to my confusion if I couldn't get velocity or time
Also, when pulled back, wouldn't there still be a height since I don't know if I'm pulling back the elastic all the way to the ground (to make height zero)
CWatters
#7
Jan8-13, 04:26 AM
P: 3,135
Ok so you know that (ignoring air resistance) you get max distance when the launch angle is 45 degrees. If you use that as the launch angle and you know the distance required you can work out the launch velocity required.
CWatters
#8
Jan8-13, 04:51 AM
P: 3,135
Concentrate on the launch velocity first but..

Quote Quote by Redfire66 View Post
Also, when pulled back, wouldn't there still be a height since I don't know if I'm pulling back the elastic all the way to the ground (to make height zero)
Yes there would but what really matters in your equation is the difference between h at launch and h when armed (pulled back). Perhaps it would have been better had I rearranged your equation like this...

mg(hlaunch - hpulled back) + 1/2mv2 = 1/2kx2

or

mgΔh + 1/2mv2 = 1/2kx2

where

Δh = xsinθ
Redfire66
#9
Jan8-13, 02:14 PM
P: 5
Quote Quote by CWatters View Post
Concentrate on the launch velocity first but..



Yes there would but what really matters in your equation is the difference between h at launch and h when armed (pulled back). Perhaps it would have been better had I rearranged your equation like this...

mg(hlaunch - hpulled back) + 1/2mv2 = 1/2kx2

or

mgΔh + 1/2mv2 = 1/2kx2

where

Δh = xsinθ
Oh I see.... sort of. Also, for determine the velocity first (sorry for jumping around), how should I approach this? Would it be possible to relate the mass of the projectile and vertical plane into the horizontal? It doesn't really seem like I should do it since they are perpendicular, but I don't have any other idea since they are related in time and velocity
I already attempted with splitting up the horizontal and vertical planes, and I have a feeling that I should begin with the vertical for to solve for time, but I get stumped which formula to use or how to approach this since I don't want to neglect the ball's mass and initial height.
CWatters
#10
Jan9-13, 01:35 AM
P: 3,135
Oh I see.... sort of. Also, for determine the velocity first (sorry for jumping around), how should I approach this? Would it be possible to relate the mass of the projectile and vertical plane into the horizontal? It doesn't really seem like I should do it since they are perpendicular, but I don't have any other idea since they are related in time and velocity..
That's exactly the right approach..

I would assume a 45 degree launch angle initially then..

Using standard equations of motion write an equation for the vertical motion in terms of:

Initial vertical velocity
Acceleration
flight time

Write an equation for the horizontal motion in terms of:

Horizontal velocity
Distance
flight time

How many unknows are there? How many equations do you have?
Redfire66
#11
Jan9-13, 11:11 AM
P: 5
Quote Quote by CWatters View Post
That's exactly the right approach..

I would assume a 45 degree launch angle initially then..

Using standard equations of motion write an equation for the vertical motion in terms of:

Initial vertical velocity
Acceleration
flight time

Write an equation for the horizontal motion in terms of:

Horizontal velocity
Distance
flight time

How many unknowns are there? How many equations do you have?
I have a few equations and not a lot of unknowns; I think that if I were to write a horizontal equation and vertical equation, I would be able to relate the two together to determine time to start it off, by eliminating velocity in the equation
basically dx = vt
but since I'm in the horizontal plane I could re-write it as
dx/vcosθ = t
And if I applied it to the vertical equation
-dy = vtsinθ - 1/2gt^2
Where dy is the height of my launcher
But how would I determine the height that it reaches over the net?
That is, there is a specific distance that it must travel... so should I use the velocity to determine how long it would take to reach a specific distance to determine if it goes above the net? (even though I'm pretty sure it would either way)


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