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Tennis Ball Launcher Calcuations 
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#1
Jan713, 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 


#2
Jan713, 05:00 PM

P: 3,106




#3
Jan713, 05:12 PM

Mentor
P: 40,717

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 


#4
Jan713, 05:33 PM

P: 5

Tennis Ball Launcher Calcuations



#5
Jan713, 06:12 PM

P: 3,106

Then regarding.... 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 


#6
Jan713, 09:20 PM

P: 5

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) 


#7
Jan813, 04:26 AM

P: 3,106

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.



#8
Jan813, 04:51 AM

P: 3,106

Concentrate on the launch velocity first but..
mg(h_{launch}  h_{pulled back}) + 1/2mv^{2 }= 1/2kx^{2} or mgΔh + 1/2mv^{2} = 1/2kx^{2} where Δh = xsinθ 


#9
Jan813, 02:14 PM

P: 5

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. 


#10
Jan913, 01:35 AM

P: 3,106

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? 


#11
Jan913, 11:11 AM

P: 5

basically dx = vt but since I'm in the horizontal plane I could rewrite 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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