I assume the question is referring to a ball say that is being kicked, in this case some fairly simplistic physics determines that 45 degrees is the perfect angle, this calculation relies upon the fact that the ground is at the same height when it was kicked as when to when it comes to rest. (Delta Yx =0. Hence a level ground)
However... If you are talking about a shot put for instance there is a different story, the optimum angle for this type of throw would be around 42 degrees. This value is different from the first due to the differential in heights between Y(x0) and Y(xfinal) where X is distance, Y states the vertical plane and 0 when t=0 (initial height) and then final when Vx&Vy=0 (aka rest).
A key point to remember in the proof of this is that velocity, time and distance are all independent of mass.
Regardless of how you choose to calculate the initial velocity the explanation stated above is always true for the angle both in a vacuum and in a real world environment. You didn't ask for a scientific proof so in summary:
45 degrees for a ground object (eg. football/soccer)
42 degrees for an object thrown from torso height (eg. shotput) - Approximation
the angle that players kick in soccer the angle they swing the bat in baseball the angle they shoot in basketball soccer balls have pentagons and hexagons on them
Talk to Mrs.Hunt dude :(
You shoot a cannon with a vi of 18 m/s. You need to get it 16m and over a 10m wall. The acceleration in the x is 0 and in the y it is gravity. The question is at what angle theta can you shoot the cannon over the fence and 16 meters away. How do I find theta?
In first-angle projection, the projectors originate as if radiated from a viewer's eyeballs and shoot through the 3D object to project a 2D image onto the plane behind it. The 3D object is projected into 2D "paper" space as if you were looking at a radiograph of the object: the top view is under the front view, the right view is at the left of the front view. First-angle projection is the ISO standard and is primarily used in Europe.
If you think about this, you can figure it out. If fish are trapped in a barrel, how could they get away from a gun? It means something is ridiculously easy.
45 degrees.
Crosman doesn't give any information about distance. You will have to experiment on your own.
45 degree angle. this gives the greatest distance. Less then 45 degrees and gravity will pull it down quicker. more then 45 degrees, and it'll go higher then it goes in distance. at 45 degrees, the height and distance will be equal, giving enough height to keep it in the air longer without sacrificing distance.
An angle-shoot is a botanical term for a side shoot which grows from the main stem.
An angle shoot is a botanical term for a side shoot which grows from the main stem.
45 Degrees is exactly between 90 and 0 degrees so it is the maximim shot. If you shoot something 90 degrees it would go straight up. If you shoot somthing 0 degrees it hits the ground fastest because of gravity. But the 45 degree angle allows for the most lift and distance for an object to travel.
Water.
All paintball gun shoot around the same distance. which is about 100 yards at a 45 degree angle, but around 50 when actually shooting.
if shooting at around 285 fps I would say 150ft max
Yes, it does. I never use more then about a 10 degree offset
Depends on which bullet, and which gun. The .50 cal Browning M-2 Machine gun has an EFFECTIVE range of 2000 yards. The maximum range (how far it can shoot when elevated to a 45 degree angle) is several MILES. Most other firearms have a range much less than that.
Hello: * Yes, the angle of a catapult does affect the distance. And this applies to both changing the angle of the catapult and changing the angle of the terrain under the catapult. If you shoot the catapult at say 45 degrees, you have very good distance. If you shoot it at 30 degrees, while it may be further, it would be lower to the ground and perhaps not travel as far hitting trees and running into wind-shear. If you shoot at say 80 degrees, while the lob goes high up in the air, it won't travel very far. So your best bet, for maximum distance, is to take your catapult to the highest possible altitude, aim for 45 degrees, taking into account the wind direction and speed, and lob away. Don't hit anyone. :) Answer Actually, the maximum distance that can be achieved from a catapult is at an angle of 45 degrees. Every degree increase from 0 up to 45 approaches the maximum distance that something can be thrown. Every degree from 45 to 90 decreases the distance.