Numerical related projectile motion??
157080
This is just the regular projectile motion problem (see your text) just with an added initial velocity in the x direction (the velocity of the ship).\n\nLet's take as our positive direction the direction the gun shoots. (That is, the ship is moving in the negative direction.) Thus, the initial x component of velocity of the projectile is the x-component of the guns velocity less the velocity of the ship. (I'm going to use Vo for the initial velocity, since using "not" as a subscript confuses me.)  And I'll use A for alpha to save space.\n\nVxo = Vo cos(A)- v\nVyo = Vo sin(A)\n\nOur equations of motion are\n\nX = Xo + Vox t + 1/2 ax t^2\nY = Yo + Voy t + 1/2 ay t^2\n\nTo find the range, we want when the projectile splashes back down again, i.e. Y = 0.  If we set Xo = 0, then X will be the range, i.e. the X position when Y = 0 again.  And there is no acceleration in the x direction, the y acceleration is -g, the acceleration of gravity.  Plugging all that in, we get\n\nR = Vxo t \nR = (Vo cos(A) - V) t                     (1)\n\n0 = Vyo t - g t^2/2\nVyo t = g t^2/2\nVyo = g t/2\n2 Vyo/g = t\n2 ( Vo sin(A)) / g = t                     (2)\n\nPlugging our expression for t (Eqn. 2) into Eqn. 1 yields\n\nR = (Vo cos(A) - V)2 ( Vo sin(A)) / g \n\nwhich is the formula you gave.\n\nTo find the max range, we can use calculus.  There may be a more straightforward way if you don't know calculus, but this works for me.\n\nThe range formula is a function R of variable A.  To find the max, take the derivative and set it equal to 0.  Thus\n\ndR/dA = 2 Vo/g [sin(A)(-Vo sin(A)) + (Vo cos(A) - V)cos(A)]\n       =2 Vo/g [-Vo sin^2(A) + Vo cos^2(A) - V cos(A)]\n\nusing a trig identity\n\n       =2 Vo/g[Vo cos^2(A) - Vo + Vo cos^2(A) - V cos(A)]\n       =2 Vo/g[2Vo cos^2(A) - V cos(A) - Vo]\n\nWhat we have here is a quadratic equation with the variable cos(A).  When we set it equal to 0, in order to find the minimum, we can use the quadratic formula.\n\ncos(A) = (- (-V) + SQRT[V^2 - 4(2Vo)(-Vo)])/2 (2Vo)\n        = (V + SQRT[V^2 + 8Vo^2])/4Vo
506920
You would agree that, the projectile motion is nothing but a trajectory. It is explained as, the path traced by an object in space.\n\nTherefore when we study the motion of a rocket in space, yes we can call it a projectile.
52010
Nancy's momentum is 270 kg*m/s west.\nTonya's momentum is 210 kg*m/s 35 degrees south of west.\n\n1. Break Tonya's momentum into west and south components using trig.\n\n2. Add their westward momenta to get the new westward momentum for the two bodies.  Tonya's southward momentum becomes the southward momentum for the two bodies.\n\n3. Divide their southward momentum and westward momentum by the combined mass of 110 kg to get the southward velocity and westward velocity.\n\n4. Use Pythagoras to get the net velocity.  Use trig to get the angle.
951198
^ Parasite alert LOL! ^\n\n30 s for C to move from A to B, maybe? I'll assume that. Also, x seems to be a length, not an angle. Maybe a) is asking for the rate of change of z (called z')? We know the angular rate of C about O (y') is pi/30 (answer to b)), and triangle CAO is isosceles, so z'=-0.5*pi/30. \nFor c), circumferential velocity is pi*r/30. Angle CAO is, conveniently, 45 deg, so x' = sqrt(0.5)*circumferential velocity.\nFor d), the area of triangle OAC is .5*AO*altitude, and altitude=r*sin(y). This is a maximum when y (not x) =pi/2.\nSorry, I know you wanted  no answers, but I couldn't see how/where to stop short of most of the answers.
141479
You should do your own homework problems.  Being told answers will not help you learn.\n\nI'll try to help you understand, though.\n\nThink about this.  The tiger's speed accelerates at a rate of 9.81 m/s^2.  For constant acceleration, distance is related to time by the equation d=t*v_0+(t^2)*a where v_0 is the initial speed IN THE DIRECTION OF ACCELERATION, d is the distance traveled, a is the rate of acceleration and t is the time.  Use that to figure out how much time it takes the tiger to reach the ground.  Then multiply this amount of time by the forward speed.  This tells you how far the tiger has moved forward upon landing.\n\nFor the ball the same formula applies.  It works for the projectile too, but you must decompose the velocity into its vertical and horizontal components (hint:  one will be the sine of the angle it's fired from times the speed and the other will be the cosine of that angle times the speed).  I'm going to leave it up to you to figure out which.  This question obviously has more to do with wanting your homework answered for you than with personal curiosity, because of its specificity, and if you just didn't get the concepts it's ok.  It is not ok to submit your homework on the internet in hopes that others will do it for you.
780136
Different trajectory?
980574
Using conservation of momentum find the velocity V of bullet+pendulum after the collision:\n\nm*v = (M + v)V, V = m*v/(M + m), V = 0.014*220/(3.1 + 0.014)\n\nV = 0.989 m/s or V = 1 m/s (approx)\n\nUsing the conservation of the mechanical energy find the height that the pendulum+ bullet goes:\n\n(M +m)*V^2/2 = (M + m)*g*h, h = V^2/2*g, h = 1/2*10, h = 0.05 m.\n\nYou can determine the horizontal displacement x using the Pythagorean theorem:\n\nx = sqrt(L^2 - (L - h)^2)\n\nx = sqrt(3.5^2 - 3.45^2), x = 0.59 m
1137727
I like the joke!  To play along, I define:\nr = metric r  and \nd = metric d\n\nV = (pi/3)*(d^2)*(3r - d)\n\nThis will work for the tank being any fraction of a sphere, not just 1/2
1008231
Set the equation of motion.\nTaking upward direction as positive\ns=vt-1/2gt^2\nat maximum height s=10,v=0\ng=-g since upward direction is positive\nyou get t= sqrt(2s/g)\n=sqrt(2*10/9.8)\n=1.428 secs
794091
I'd suggest you first try the site listed below where you can find a downloadable spreadsheet that predicts velocity. You can deduce the author's equations from it. It's actually a time-stepping simulation rather than analytical equations. However, a bit of theory wouldn't hurt either. First, to keep it simple, let's assume the pressure is constant for T, the lesser of the time the valve is open and the time the projectile remains in the barrel. This is actually not a good assumption since many spudguns have pressure vessels that are smaller than or not much bigger than the barrel volume. Anyway, during T the spud is subjected to a force F=P*barrel area = P*pi*d^2/4. The net force accelerating the spud is F - friction force, so its acceleration A=(F-Ffriction)/M and its velocity v=A*T. I don't know the practice on valving. It may be that open valve time is always less than barrel transit time, but if not, you can compute the distance S that the spud would travel during the open valve interval T; S=0.5*A*T^2. If S>l, then T is the time spent in the barrel, which is SQRT(2*l/A), and you should recompute v. \nAnother assumption would be that each shot entirely empties the pressure tank. In this case we simplify by disregarding pipe friction, thermodynamic losses (like failure to fully expand due to cooling) and any other losses you can think of, and assume the potential energy of the compressed air is converted to kinetic energy. The potential energy PE = 0.5*P*vol (where vol is tank volume = L*pi*D^2/4). The kinetic energy of the spud is 0.5*M*v^2. Thus v=SQRT(P*vol/M).\nAbout other questions, 1 psi = 6894.75729 Pascals, and speed of sound =340.29 m/sec.\nNote that with what you have given in your question, you have a lot of units conversion to do.\nEDIT: See friction stuff added above. The spreadsheet includes friction effects.
551475
You could start by calculating the initial velocity and launch angle required for each distance and then design your catapult to meet those requirements. The below simulator can help with making the velocity and angle determinations.\n\nhttp://www.lcse.umn.edu/specs/labs/catapult/practice.html\n\n\nPersonally I would try to get an idea of what I needed to build from calculations. Then build something that I thought was over kill, and include adjust ability. Then I would figure out how far back the arm had to be pulled to give me the desired distances by trial and error. Getting the calculations to match real world conditions could get complicated.
479843
I'm assuming it's okay to neglect air resistance in this problem, and that the vertocal velocity component you requested is right before the thingy (I love the word "thingy") hits the ground.\n\nWe have the kinematic equation: S=160+V(initial)t+.5at^2.\na is -32 ft/sec^2, and the initial vertical velocity is 0, so we have S=160-16t^2.  When it hits, S=0, so:\n\n160-16t^2=0 --> 16t^2=160  -->  t^2=10  -->t=10^(1/2)\n\n(Or t = (square root of ten) seconds.  About 3.16 seconds.)\n\nNow that we have t, we just multiply that by the horizontal velocity:\n\n960*10^(1/2), or 960 square root of 10.  (In feet.)  \n\nTherefore the object will land 960*10^(1/2) feet from the base of the building.  Roughly this is about 3035.78 feet.\n\nTo get the final vertical velocity, we use the equation:\n\nV(final)^2=V(initial)^2+2a delta S.  \n\nChange in S in the vertical direction is -160 ft, and a=-32 ft/second squared, and V(initial)=0, so:\n\nV^2=2(-32)(-160)=10240 ft squared per second squared.\n\n--> V=32*10^(1/2), or about 101.19 feet per second.\n\nI hope that's correct.  I typed it with one hand while not really paying attention.  ^_^  From now on, do your own homework.  :P
916863
I get that the maximum distance that the ship can fire is 6371m, meaning that any distance greater than 3571m from the western shore should be safe.  It may be tempting to assume that the target ship would be safe within a certain minimum distance, where it would be protected by the mountain peak, but the enemy ship could hit any target within the 3571m limit by angling its guns higher than 45 degrees.\n\nEDIT:\nTo start solving our problem, let's figure out the maximum distance that the enemy ship can fire.  It's not hard, and we know we'll have to figure it out anyway.\n\nIf the ship fires its projectile at exactly 45 degrees, the projectile will make a parabolic arc through the air.  At angles of more than 45 degrees, the projectile will fly higher, but it won't go as far (like a pop-up fly in baseball).  At angles less than 45 degrees, the projectile will fly lower, and hit the water with more horizontal motion (like a line-drive in baseball).  At exactly 45 degrees, it will achieve its maximum horizontal distance.  \n\nIn order to figure out how far the projectile flies, we need to break up its motion into horizontal and vertical components.  Imagine a right triangle with a hypotenuse of 250 units and one angle of 45 degrees.  (The other angle will also be 45 degrees, but that doesn't matter for now.)  The line adjacent to our angle will be as long as the projectile's horizontal momentum.  The line opposite the angle will be as long as the projectile's vertical momentum.  We just need a little bit of trigonometry to figure out these numbers.\n\nsin(a) = opposite / hypotenuse;\nsin(45) = Vv / 250;   (Vv represents the vertical velocity)\nVv = sin(45) * 250 = 177m/s;\n\nSince we're using a right triangle with two 45 degree angles, both sides will be the same length.  If we were using a different angle, we could easily solve for the horizontal momentum using the Pythagorean Theorum.\n\nSo, we know that our projectile had a horizontal velocity of 177m/s and a vertical velocity of 177m/s.  Let's figure out how long it would stay in the air.  We know that it's constantly accelerating towards the earth due to the force of gravity.  Our answer lies in the definition of acceleration: acceleration is a change in velocity over time.  Mathematically, that's (a=ΔV / t).  The "Δ" symbol is the Greek letter delta, often used to represent a change in a value.  \n\nTo figure out how long it stayed up, we only need to worry about the projectile's vertical velocity.  We know that it's initial vertical velocity is 177m/s.  We also know that it's vertical velocity at the top of the arc is 0m/s.  The change between these two points in time is (177m/s - 0m/s =) 177m/s.  The only other thing we need to figure out time is the acceleration due to gravity.  On Earth, it's always 9.81m/s^2. So...\n\na = ΔV / t;\n9.81 = 177 / t;\nt = 177 / 9.81 = 18.04 seconds.  That's how long it takes the projectile to reach the top of it's arc.  It will take equally as long to reach the bottom of it's arc, so the total flight time is 36.09 seconds.  \n\nNow we have a simple distance equation.  Since distance is equal to velocity multiplied by time, the projectile's horizontal distance will be equal to its horizontal velocity multiplied by the amount of time that it stayed in the air.\n\nd = 177 * 36 = 6372m.  (This number is a little higher than the first one, because I rounded a little differently.  It doesn't make a huge difference.)  If we take out the distance to the western shore (2500m + 300M), we get a maximum distance of 3572m from the western shore.  Any ship beyond that distance will be safe.  \n\nAll we need to do now is figure out whether or not the enemy ship can fire at that maximum distance.  If the mountain is too high, the ship will have to increase the angle of its guns, thereby losing range.  We can figure out how high our projectile is going to fly with
81906
First Question:\nVertical and horizontal motion are independent from each other.  Use d=(v.i)(t)+(1/2)(a)(t)^2 to find the time that the dart is in the air.  d is distance(0.32m), v.i is the initial velocity(0.00m/s vertically since it was thrown exactly horizontally), a is acceleration (9.8m/s^2: gravity) and t is the time in seconds.  Now that you have the time, use that in the equation x=vt, where x is the horizontal distance (range) that you are asked to solve for, v is the velocity(11.8m/s), and t is the time that you just found.  The answer to the problem is the x variable in this equation.\n\nSecond Question:\nI have attempted this problem for about 40 minutes and I have tried every possible combination of variables and equations, but there is not enough information to solve it.  Therefore, I have to say that it is an impossible problem.  If you figure out how to solve it, please, send me an e-mail at the address I give later and tell me how to do it.\n\nThird Question:\nFirst, find the distance that the basket is from the launch point by subtracting the two.  Then, use (v.f)^2=(v.i)^2+(2)(a)(d), where v.f is the final velocity, v.i is the initial velocity, a is the acceleration (9.8m/s^2), and d is the distance which you just found.  To  get the ball to the basket, it only needs to go the exact distance to the basket, so the final velocity is 0.00m/s.  Use the previous equation to find the intial vertical velocity.  The use v.iy=(v)sin(angle), where v.iy is the initial vertical velocity and v is the initial angular velocity, to find the velocity the ball must be launched at for each angle.\n\nPlease send me an e-mail at bge1234@yahoo.com after you turn in your work and get the solution so that I can see how your teacher did it.  In particular, tell me how to do the second question.  My ways SHOULD work as long as you don't make any calculatory errors.  I hope this helps you.
438060
Without regard to air resistance you solve the problem as such.\n\nv(f) = final velocity\nv(i) = initial velocity\na = acceleration\nt = time\nx = displacement (distance)\n\nWhen you throw the ball up vertically, you know several things:\n1) v(i) = 30 m/s\n2) v(f) = 0 m/s\n3) a = -9.8 m/s^2 (this is because acceleration due to gravity goes in the opposite direction of the object)\n\nSolve for x with this equation:\nv(f)^2 = v(i)^2 + 2ax\n0^2 = (30 m/s)^2 + 2(-9.8 m/s^2)x\n0 = 900 m^2/s^2 - (19.6 m/s^2)x\n(19.6 m/s^2)x = 900 m^2/s^2\nx = (900 m^2/s^2)/(19.6 m/s^2)\nx = 45.9 m (rounded off)\n\nNow you know how high the ball went.  The time it takes to reach this is equal to the time it takes to get back down (assuming that it is the same distance up and down).\n\nTo solve for t, use this equation:\n\nv(f) = v(i) + at\n\n0 = 30 + (-9.8)t\n9.8t = 30\nt = 30/9.8\nt = 3.06 s  this is going up to max height.\n\nto check if it will be the same time going down, use this equation:\n\nx = v(i)t + (1/2)at^2\nv(i) = 0 (because when you reached max height, your velocity will be zero)\na = 9.8 because your ball is travelling in the direction of the acceleration.\n45.9 = 0 + (1/2)(9.8)t^2\nt^2 = 45.9/4.9 = 9.367\nt = 3.06 s\n\ndone
397135
The weight of the object is irrelevant unless one is taking into account air resistance. This is because the net force on the object is the force of gravity, so we have ma = -mg, or a = -g, which is all we need to know for the vertical kinematics equations. The only thing we need to know for the horizontal equation is the initial horizontal speed of the package.\nIf, on the other hand, one was concerned with air resistance, one would also need to know approximate air density and cross-sectional area of the object, as well as be able to numerically solve a nonlinear ordinary differential equation (the frictional force is proportional to some function of velocity) so I doubt this is what you wanted.
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)