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A horizontal rifle is fired at a bull’s-eye. The muzzle speed of the bullet is 670 m/s. The gun is pointed directly at the center of the bull’s-eye, but the bullet strikes the target 0.025 m below the center. What is the horizontal distance between the end of the rifle and the bull’s-eye?
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Physics 101 Mechanics
Kinematics in Two Dimensions
Motion in 2d or 3d
University of Washington
Simon Fraser University
2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.
A vector is a mathematical entity that has a magnitude (or length) and direction. The vector is represented by a line segment with a definite beginning, direction, and magnitude. Vectors are added by adding their respective components, and multiplied by a scalar (or a number) to scale the vector.
A horizontal rifle is fire…
Bull’s-eye Bob at a huntin…
A bullet is fired from a r…
A rifle that shoots bullet…
A rifle has muzzle velocit…
A person holds a rifle hor…
A rifle is aimed horizonta…
All right. So here, we're gonna problem. We're, uh 18. Ah, First things first. With this problem, we want to find the time that the bullets in the air and wouldn't do that. Using the formula distance in the Y is equal to the initial velocity in the wild times a time plus 1/2 acceleration times squared. What? We know this isn't traveling anywhere in the wind direction. It's not going well. Initially, I guess it's not going anyway in the wind direction. So that term is just zero and a Here is acceleration due to gravity. So we can write distance in the wind is actually equal to 1/2 g t squared because acceleration is equally gravity rightto factors on. Then we want Thio said that we wanna solve for times we're gonna have t squared is equal to two d sub y over G Uh, so then we can just, uh, take that over there and just put t and square root This that's really big. I guess that make a difference. So equals two times 0.0 to 5 meters, divided by gravity, which is 9.80 meters per second squared scared That gives us a time of 7.143 times 10 to the negative two seconds. Okay, so now we have a time of the boards in the air. So now we confined our how far it goes in the extraction. So start anyway. Word. So, uh, time was 7.1 for three times 10 to the negative, two seconds. So now that she's the same formula that we did for why but in the ex direction so d X equals V initial blasting. Next times time 1/2 a t squared. But we're not accelerating in the axe. We're ignoring a resistance, so we have a constant velocity. So with no acceleration, this goes to zero and so pretty much we have distance in the X is equal to V I X times time, its initial velocity in the extremes that time so we can plug in the numbers that we had from the questions we had 680 meters per second and then we can play in this time appear that we found so 7.143 times 10 to the negative two seconds gives us ah ex distance of 48 meters, So, yeah, I always find the time first using one of the forms that you have. Uh, the rest is pretty simple from there. Just plugging Chuck.
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