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Trebuchet Projectile Physics  Height Formula 
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#1
May2808, 02:17 AM

P: 7

Hey guys, this is my first post, so im sorry if this is in the wrong section. For physics at school, i have built a trebuchet, which stands about 1.2 meters tall, and the arm is about 2 meters long. I have it throwing a projectile easily enough, but for my experiment, i need to find out the following data;
1) Displacement of Projectile This is easy enough, simply measure the distance from where the projectile takes off, and where it lands. 2) The average velocity This is also easy, by using the displacement, and the time taken to get there. 3) The largest height I have not figured out an accurate way to do this yet, but please correct me if im wrong when i say this. Shouldn't I be able to find out the largest height of the projectile, by using the time taken, and the displacement? For example, if in test one the projectile go 20m long, and it takes 5 seconds. Then if in test 2 it does the same length, except it takes 6 seconds, this will mean that it went heigher. Anyway, now my question is, what is the formula for working out this? 4) The distance of the projectiles flight path This one is tricky, i have spoken to my teachers about this and they say that it's not expected of me to do this, because some of them didn't even know how i would do this. I came up with a solution which i think is very close to correct, however it will be a little out. But it will still be close enough. If i get the arc of the flight path (in graphmatica), by adding in the starting point, heighest point and landing point, i should then be able to draw many vertical lines through this flight path, and then some horozontal lines through those vertical one's, where the vertical one's meat the flight path. So it should look like this; //img79.imageshack.us/img79/1197/physicsformulasz6.png (add h t t p : / / in front of it. I couldn't do it because it's my first post, and you cant post URLS on your first post) Now, see how i've made the circle around one point, and notice how each of these areas are now close to a right angled triangle. Well i was thinking, i could get the hypotenuse of each of these little triangles, and add them all up, and that would get me the distance of the flight path (Close to it, a little shorter). However, that is a lot of stuffing around. Is there a formula that can work this out, if i already new the time taken, the displacement, and height? Thankyou very much for your time. I really hope you can answer my last 2 questions! Thanks, Scotty 


#2
May2808, 03:07 AM

Sci Advisor
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P: 8,953

The maximum height can be calculated from the distance/time quite easily if you ignore air resistance  the problem is normally in most textbooks as a cannon ball.
The total path requires calculus, but your idea of using small triangles is correct. Calculus is just a mathematical version of small triangles! 


#3
May2808, 05:39 AM

P: 7

Thankyou very much, so could you please provide the height formula? Thanks, Scotty 


#4
May2808, 05:47 AM

Sci Advisor
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P: 8,953

Trebuchet Projectile Physics  Height Formula
Some good descriptions of the solution here (and the other lessons in the chapter)
http://www.glenbrook.k12.il.us/gbssc...rs/u3l2c2.html 


#5
May2808, 06:24 AM

P: 7

Thanks, however the formula which im after involves 'vertical velocity'. Im not sure how to get the vertical velocity, as some of the examples had a vertical and horizontal velocity which were different. Sorry, im not a scientist/physics person. Here's whats going to happen. I'll be doing tests with my trebuchet, and i'll have to find out all of these things from scratch. So i'll automatically be able to get the time of flight, and displacement. I can get the distance of the flight path if i work out the height, but the formula on that link uses vertical velocity. Would just using the average velocity work? (distance/time)
For example, i shoot the projectile 20 meters (displacement) It took 5 seconds for it to land. From that, the average velocity is 4m/s. (20/5) Now, to get the height, i use this formula; y=Vertical Velocity * time + 0.5 * 9.8 * time2 (time squared) so, y=4 * 5 +0.5 * 9.8 * 5(squared) so, y=102.5m (this is the height) That doesn't look right, could someone please help out here? Instead of taking me to other links, please explain to me what im doing wrong, and please provide an example with working out etc. Thankyou! Scotty 


#6
May2808, 07:20 AM

Sci Advisor
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P: 8,953

You would use the equation s = ut + 1/2 at^2
Where s is the distance, u is the initial velocity, a = 9.8m/s^2 and t is time. But you don't know the vertical velocity! So the trick is to consider the second half of the path as the projectile comes back down. Only thinking about the vertical part of the motion, at the top of the curve the projectile for an instant has no vertical velocity  as it goes from going up to coming down! So it is exactly the same as if we had simply dropped a weight from the top of the path. We know the time to fall  since it is simply half the time for the total flight, and we know 'a'=g so, s = 0*t + 1/2 * 9.8 * t^2 = 9.8 * 2.5^2 = 61m It's a little more complicated for a real trebuchet because the projectile is released a distance above the ground  but this is close. 


#7
May2908, 04:32 AM

P: 7

Sorry, by 'Include an example' i meant include an example with an experiement, for example, we know the displacement of the shot (starting point  ending point), and we know the time taken, then show the formula for the maximum height. This would be easier, please convert your formula to get the maximum height for the following details (this is just one last example, so i fully understand)
Displacement=40m Time Taken=5 Seconds Maximum Height=??? Thanks, and sorry for being so dumb (im not the best at physics!!) Scotty 


#8
May2908, 10:35 AM

P: 33

the key to this problem is the angle. if you want the most accurate height, get the angle from the base to the arm. see attatchemnt for explenation. (the picture may take a second to load)



#9
May2908, 08:23 PM

P: 33

the important thing you have to remember is that distance/time isnt the initial vertical velocity. that is the the initial velocity itself. you have to find the x and ycomponants of that initial velocity. the two componants are independant of each other. the x componant is constant through out the flight, assuming no air resistance or interferance. while the y continuously changes.



#10
May3008, 06:12 AM

P: 7

Thanks, but how would i find the angle? Each problem is just causing another problem. From some research, here's what i've got.
For Maximum height point on graph x=(gravity * time (squared))/2 y= v (squared) / 2 times Gravity I think the 'v' is the vertical velocity, and that's the only variable that i wouldn't have with these formulas. So i need to find out what the vertical velocity is. BUT to do that, i need to use the angle or the arm when the projectile is released, correct? If this is the case, what would be the best way to work out the angle which the projectile was released at? Assuming that the flight path is parabolic (I think i spelt it right), shouldn't i be able to work out the angle, just by the time taken in flight, the horizontal displacement, and the overall average velocity? (d/t). As in, if the displacement is 20m, and the time taken is 5 seconds, that means that the average velocity is 4m/s. From this, shoudn't i be able to work out the angle that the projectile was launched at? Thanks, and again, sorry for taking so long to understand. It may be simple to you, but this is new to me (treat me like a new student please, i need explanations, examples..) Thanks for your time, Scotty 


#11
May3008, 11:19 AM

P: 33

becareful when you say average velocity. that is the average velocity, but only for the X componant. that is not the average velocity for the Y.
see new attatchment again for explenation on how to find the angle. 


#12
May3108, 07:43 PM

P: 7

'Attachments Pending Approval'  That message has been there for 2 days!! I can't see the document yet.
Scotty 


#13
May3108, 08:22 PM

P: 33

thats odd... i dont know why its doing that. ill try posting it again. it didnt do that with the last one i dont think.



#14
Jun108, 03:46 AM

P: 7

Ok so this is what i need to do, to find the maximum height of the projectile in it's flight path;
1) Work out the angle from the base of the trebuchet, to the point where the ball is released, by doing tan1(y/x). y is the height from the base of the trebuchet, to the top of the arm, where the projectile is released. x is the width of the base. 2) Use this formula; height = [(tanθ)(X/t)]^2 + (sinθ)(2m) 2g The angle is the angle that we've just found. X, is the distance that the projectile was thrown. t is the time that the projectile took to land. g is gravity (9.8m/s). Is that correct? Thanks, Scotty 


#15
Jun108, 02:09 PM

P: 33

That is what i got yes. good luck!



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