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 A bullet of mass 30g is fired horizontally from a rifle situated at the top of a cliff of heaight 200 m. How far from the base of the cliff will the bullet land if it's muzzle velocity is 400ms^-1. Using projectile motion formulas i got 16 000 km or something near that. >_< Help would be appreciated, working out included preferably. [Quote] [Link]
 m = 0.030kg Δy = 200m ux = 400m/s vy = uy + 2aΔy vy = 0 + (-9.81)*(2)*(200) vy = 3924m/s vy = uy + at - 3294 = 0 + (-9.81)*t t = 400s Δx = uxt Δx = (400)*(400) Δx = 160000m Δx = 160km EDIT: Oh, I see what I did wrong. vy2 = uy2 + 2aΔy vy2 = 0 + (-9.81)*(2)*(200) vy = 62.64m/s vy = uy + at - 62.64 = 0 + (-9.81)*t t = 6.3855s Δx = uxt Δx = (400)*(6.3855) Δx = 2554m Yep, that's better. [Quote] [Link]
 o.o So the bullet travelled 160 km? Edit: and it took it 400 seconds to land? Doesn't sound right to me.. but thanks. [Quote] [Link]
 Blake Webmaster2082 Posts11 Vigeo 5:5 - 3.4.63774.543 days ago Figure out the y component first. The bullet is just in free fall with an initial y-velocity of 0. d = 1/2 * a * t2 d = 400, a = 9.81, solve for t t = bullet was in the air for 6.3855 seconds before it hit the ground. Figure out the x component. Bullet will keep travelling at 400 m/s sideways until it hits the ground. 6.3855 * 400 m/s = 2.554 km from the cliff. Mass of the bullet is irrelevant. [Quote] [Link]
 No disrespect to hobbaloo, but go with Blake on this one. He is correct. [Quote] [Link]
 Yyyeah, not sure exactly what went wrong there... 160km is suspicious. Oh well, no harm in trying. [Quote] [Link]
 I know a set of bugs will eat away at land and will double the area they consume each year. I also know they finish eating all the land in 10 years. However, how fast will they consume the land twice as many bugs to begin with? [Quote] [Link]
 Assuming the bugs eat at a constant rate - 9 years, of course. On the bullet problem - it will be a wee bit nearer if you factor in air resistance, and a smidgin further if you take the curvature of the Earth into account. Otherwise, Blake is spot-on as usual. [Quote] [Link]
 Well, if they consume 1 unit the first year, they consume a total of (1 + 2 + 4 + ... 512 = ) 1023 units of land. So, if they instead start at 2 units of land, at the end of year 9 they've consumed (2 + 4 + ... 512 = ) 1022 units, leaving 1 unit of land left to be eaten. So it will take 9 years, plus 1/1024 of the tenth year, so 9.00097656 years. (This is obviously assuming they double at the end of each year, and have a constant feeding rate during the year). [Quote] [Link]
 I believe that you are correct with one small error. They should consume 1/1023 on the final year, not 1/1024. That being said, the margin of error is already extremely tiny, so I personally find it arbitrary. [Quote] [Link]
 More of a physics problem, but I figure it can fit here. A duck has a mass of 2.5kg. As the duck paddles, a force of .10N acts on it due east. The water current exerts a force of .20N 53 degrees south of east. When these forces begin to act, the velocity of the duck is .11 m/s due east. Find the magnitude and direction (relative to east) of the displacement of the duck after 3.0 seconds. I've worked through this one twice already, and I've gotten it wrong each time. EDIT: Vector drawings would be really cool, too. [Quote] [Link]
 First, you must break up the south-of-east force into its components. You can do this using trig. x-component=.2Cos(53)=.1204 y-component=.2Sin(53)=.1597 From here you add the x-componet of the south-of-east force to the eastern force to get a force in the x direction of 1.1204. Because F=ma, the acceleration in a given direction is F/m. So, the acceleration in the x and y directions are as follows: x-acceleration=1.1204/2.5=.4482 y-acceleration=.1597/2.5=.06389 You then plug this into your postion functions (a/2)t^2 + vt + x to determine position: x-position: .4482(3)^2 + .11(3) = 4.3638 y-position: .06389(3)^2 = .575 From here, you use the pythagorean theorem to determine magnitude of displacement: Sqrt[4.3638^2 + .575^2] = 4.4015 meters You can then use any number of various trig operations to find the angle: theta = 7.506 degrees south of east I'm sorry I couldn't give you any vector diagrams. Hopefully you can figure it out. (Double check the math as well). [Quote] [Link]
 larana said:I know a set of bugs will eat away at land and will double the area they consume each year. I also know they finish eating all the land in 10 years. However, how fast will they consume the land twice as many bugs to begin with? My friends showed this to my math teacher, she had them do it on the board, I was stumped until I saw how they did it, they got aroud nine years something (Like many have said) [Quote] [Link]
 Easy question here: For all n, prove that 1^n + 8^n - 3^n - 6^n is divisible by 10. [Quote] [Link]
 Blake Webmaster2082 Posts11 Vigeo 7:2 - 1.73.26765.609 days ago To check for divisibility by 10, we only need to know what the last digit of each of the terms of the summation will be. The last digit of the powers of 8 repeat in a sequence of 4: 8 -> 64 -> 512 -> 4096 -> 32768. Simple rules of arithmetic make it clear that the ones digit of the product of two numbers can only be influenced by the ones digits of the two factors being multiplied together. Therefore this pattern is guaranteed to repeat. Similarly the powers of 3 do the same. Its ones digit sequence is 3 -> 9 -> 7 -> 1 -> 3 which is also a sequence of length 4. 6's raised to any positive power will always end with a 6. Because the powers of 3 and 8 repeat in a sequence of 4 and the powers of 1 and 6 are always 1 and 6, we need only to check to see if the given holds true for n between 1 and 4. Additionally, we don't even need to figure out what the powers are. Simply the last digit from the repeating sequence will do. n = 1: 1 + 8 - 3 - 6 = 0 (yup) n = 2: 1 + 4 - 9 - 6 = -10 (yup) n = 3: 1 + 2 - 7 - 6 = -10 (yup) n = 4: 1 + 6 - 1 - 6 = 0 (yup) n = 5: 1 + 8 - 3 - 6 = 0 (now it's repeating) [Quote] [Link]
 I did it slightly differently. Looking in Modulo 5, the equation is 1^n + 3^n - 3^n - 1^n, giving a remainder of 0 Looking in Modulo 2, the equation is 1^n + 0^n - 1^n - 0^n, again giving a 0 remainder. Since the result is divisible by both 5 and 2, it must be divisible by 10. One caveat - "for all n". It only works for natural numbers. Using -1 quickly gives a fail. [Quote] [Link]
 DIAV said:I did it slightly differently. Looking in Modulo 5, the equation is 1^n + 3^n - 3^n - 1^n, giving a remainder of 0 Looking in Modulo 2, the equation is 1^n + 0^n - 1^n - 0^n, again giving a 0 remainder. Since the result is divisible by both 5 and 2, it must be divisible by 10. One caveat - "for all n". It only works for natural numbers. Using -1 quickly gives a fail. Ah, must have written the question wrong, then. I got it from my school's math club. [Quote] [Link]
 I need to change this problem to standard form (ax2 + bx + c = 0) and factor it. Problem is, I don't know how to get it to standard form. So, if someone could help that would be awesome. Thanks! [Quote] [Link]
 To get it into standard form, multiply both sides by two. This gives you: 2x^2 - 2x = 12. From there, subtract 12 from both sides: 2x^2 - 2x - 12 = 0  I fixed my mistake. [Quote] [Link]