#1
How do i prove that the optimum angle for firing a projectile is 45 degrees?
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#3
I was unaware that shooting something at 45 degrees is best, I'll keep that in mind the next time I play BowMan on addictinggames.com...
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#4
45 degrees goes the farthest.

u like... derive the kinematic equations. the 1/2*g*t^2 one for the vertical (y) position, and then Vf = Vi + at or sumtin like that with the horizontal (x) position.
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#5
Quote by i(ncubus)play
I was unaware that shooting something at 45 degrees is best, I'll keep that in mind the next time I play BowMan on addictinggames.com...



Lmao.


Anyway. It is the two vectors, duh.
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#6
the reason you split it into its vectors is because it shows that you get the greatest ratio of your x vector to your y vector, that ratio being 1. your y vector represents time spent in the air, and x vector represents x-ward velocity
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#8
you could also run tests. Firing an object at a 45 degree angle, then firing the same object from a 55 degree and a 35 degree angle (or whatever angles you want to use).
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#9
Quote by eviledge87
you could also run tests. Firing an object at a 45 degree angle, then firing the same object from a 55 degree and a 35 degree angle (or whatever angles you want to use).

thats what our experiment was. We had this projectile and a small cannon and had to change the incline by 10 degrees each time up to 70 degrees. The trouble is, we didn't have any masses, forces or velocities to work out the scientific way. Is there some theory as to why it is 45 degrees?
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#10
Unless you're actually going to try and shoot something really far.

In which case 45 degrees is bad, cuz it'd like... I ono... hit something.
#11
write of the formula for the distance of the projectile in relation the the angle which it is thrown at

differentiate said formula
#12
Use the range formula.

range = v0^2 * sin (2 * theta) / g

Take its derivative, set it equal to zero, and solve for theta. You'll get 45 degrees (or pi / 4 radians)
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#13
Because

Sin45 = 1/2 = X Vector
Cos45 = 1/2 = Y Vector

Therefore the X direction and the Y direction are at their best potential.
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#14
Quote by aprescott_27
Use the range formula.

range = v0^2 * sin (2 * theta) / g

Take its derivative, set it equal to zero, and solve for theta. You'll get 45 degrees (or pi / 4 radians)

range = v0^2 * sin (2 * theta) / g
drange/dtheta = v0^2/g * cos (2 * theta)*2
0 = v0^2/g * cos (2 * theta)*2
0 = cos (2 * theta)
2*theta = pi/2 (+k*pi if you wish)
theta = pi/4 = 90°
#15
Quote by seljer
range = v0^2 * sin (2 * theta) / g
drange/dtheta = v0^2/g * cos (2 * theta)*2
0 = v0^2/g * cos (2 * theta)*2
0 = cos (2 * theta)
2*theta = pi/2 (+k*pi if you wish)
theta = pi/4 = 90°

pi/4 = 45°
Last edited by rayIII at Oct 31, 2007,
#16
Quote by rayIII
pi/4 = 45°


yeah, its 2am so i'm not thinking

I guess they really have managed to cram radians into my head more than degrees
#17
Quote by seljer
yeah, its 2am so i'm not thinking

lol nice at least tonight you can blame halloween
#18
Quote by rayIII
lol nice at least tonight you can blame halloween


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