Is there an optimal angle for shooting projectiles as far as possible?

Question

Consider a cannon firing something into the air. If the objective was to launch the shot as far as possible, what angle would the bore have to be in relation to the ground? If all factors were constant, is there a specific angle for max distance? One that is not too dramatic nor too shallow, to get the projectile as far from it's starting position as possible?

Update:

lengthy details are good

jcherry_99 · Accepted Answer

Details you want, details you get.
x = optimum angle
v = initial velocity delivered at some angle

Vertical expression
vi = v * sin(x)
g = -9.8  m/s^2 the minus is because I've arbitrarily chosen up is as plus. Down is therefore -.
vf = 0
vf^2 = vi^2 + 2*ad
a = (vf - vi) / t
t = (0 -  v * sin(x) ) / -9.8
t = v * sin(x) / 9.8
This gets you to the maximum height, but it does not represent the total time. You have to multiply by 2.

t = 2*v * sin(x)/9.8

Horizontal expression
The acceleration in the horizontal direction is 0 m/s^2
d = v*cos(x) * t [ substitute t into this equation ] 
d_max = v^2 * 2 * sin(x) * cos(x) 
but 2 sin(x) * cos(x) = sin(2x)

d_max = v^2 * sin(2x)

And now for the stinker.  d will be a maximum when the differential of d = 0

d_max/dx = v^2 * (2) * cos(2x) = 0 
divide by v^2 * 2
cos(2x) = 0
2x = cos-1 ( 0 )
2x = 90o
x = 45o

Lengthy and ugly but true.

? · Answer

yeah it is 45 degree from horizontal and there is a proof to it but it is little lengthy and requires a background knowledge of projectile motion.if you want that then let me know through this page.

? · Answer

45degrees above the horizontal, or pi/4 radians above the horizontal...that's in the absence of air friction.  With air friction, it varies (depending on the air friction)

? · Answer

Six degrees from destination

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Is there an optimal angle for shooting projectiles as far as possible?

4 Answers