Projectile trajectory - Trebuchet

AI Thread Summary
The discussion revolves around calculating the velocity of a projectile launched from a trebuchet, where initial calculations yielded an implausible velocity of approximately 277 m/s. The participants used a formula for projectile motion but struggled with the results, particularly since the projectile only traveled 20 meters. A recalculated velocity of 40.5 m/s was obtained, but this still seemed high. Alternative energy-based calculations suggested a velocity of 18.89 m/s, which appeared more reasonable, though there was uncertainty about the methodology used. The conversation highlights the complexities of applying physics equations accurately in practical scenarios.
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Homework Statement



Hello, I have an issue regarding projectile trajectory.
As an assigment we have made a trebuchet, which is basically a catapult with a sling attached to it.

We have done some measurements with the trebuchet itself, and have actually achievement some (according to us) correct results. The theory however, gave us a strange result.

We decided we wanted to know the velocity of our projectile (which should have been easy..)

Which we determined ourself, and confirmed using wikipedia.

The velocity we received from calculating this ( see 3 ) was approx 277 m/s. Which of course, is impossible since our projectile traveled only 20 meters and did not reach it's destination that fast, what did we do wrong?

We assumed we shot approx. at 45 degrees, making alpha = 45.
The distance traveled was 20 meters, making d = 20.
The g is the gravitational force, since we were testing on our own planet g = 9.81.
y_0 is the height we shot the projectile from, in our case this was 80 cm, making y_0 = 0.80.

Homework Equations



We used this formula:
d = v*cos(alpha)/g * ( v*sin (alpha) + sqrt((v*sin(alpha))^2+2*g*y_0) )

The Attempt at a Solution



Using the above variables.. we calculated the velocity like this:

20 = v*cos(45)/9.81 * ( v*sin ( 45 ) + sqrt((v*sin(45))^2+2*9.81*0.8) )
20 = v*cos(45)*v*sin(45)/9.81 + v*cos(45)sqrt((v*sin(45))^2+2*9.81*0.8)/9.81

Where..
cos(45) = sin(45) = .5sqrt(2)
This gives us:

20 = 0.5*v^2/9.81 + .5*v*sqrt(2)*sqrt(.5*v^2+15,696)/9.81

20^2 = (0.5*v^2/9.81)^2 + (0.5*v*sqrt(2)*sqrt(.5*v^2+15,696)/9.81)^2
400 = .25*v^4/9.81^2 + .5*v^2*(.5*v^2+15.696)/9.81^2
400 = .5*v^4+7.848v^2/9.81^2

.5*v^4 + 7.848*v^2 = 38494.44
v^4+15.696v^2 = 76988.88
v^2(v^2+15.696) = 76988.88

v^2 = 76988.88 OR v^2 + 15.696 = 76988.88
v = 277.47 OR v^2 = 76973.184

v = 277.47 OR v = 277.44Thanks in Advance for checking out my problem.

p.s. I just figured this might have fitted better in the Advanced Physics part?
 
Last edited:
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Okay, so I took my calculator, and used it to calculate the answer instead..
This however, gave me a different answer:
v = 40,5 m/s

This still seems a bit high, and strangly I can't figure out what I did wrong.

As a different approach, I tried calculating it without the distance (since this would also be part of the theory of the assigment)

The variables used here are:
mass counterweight (gravitational energy) = 2.0 kg
mass projectile (kinetic energy) = 11 grams = 0.011 kg

Distance point of rotation -> counterweight = 0.25 m
Distance point of rotation -> projectile = 0.75m + 0.5m sling = 1.25 m
Distance the counterweight falls before the projectile is released: approx. 0.5m

Equations
Egrav = mgh
Ekin = .5mv^2

M = F*r (??)

Attempts..
Egrav = 2.0 * 9.81 * 0.5
Egrav = 9.81 Joule

Now, here is the problem, I am not sure if I am allowed to do this:
Ekin*1.25 = Egrav*0.25
Ekin = 1.962 J

1.962 = 0.5 * 0.011 * v^2
v = 18.89 m/s

This however, seems like an answer that can very well be correct - I am just not certain if this is a correct way of calculating it.
 
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