What is the relationship between Angle of inclined plane and velocity of object (ball)? Is it linear or power?
961575
I may be wrong but I think the relationship is uncle and nephew.\n\nGo on - report me.\n"Please sir, Slartibartfast is not playing nicely . . . . . ."
853525
Both the above answers are correct, provided the acceleration is horizontal. If it's not then it's a vectors and trig. question.
168069
assuming your statement is true, then\na=k/t\nwhich gives t=k/a\nso time is a nonlinear function of acceleration.\nGraphically, the plot of a versus t is not a straight line (it is a hyperbola)
1099642
You should probably skip what I wrote and go straight to the link http://whatis.techtarget.com/definition/0,,sid9_gci529957,00.html :-)\n\nYou measure the distance fallen in meters.  Let us say that this distance as a function of time t in seconds since the ball was dropped in a vacuum is s(t) meters.\n\nThe "instantaneous" downward velocity of the ball at time t is just the derivative of s(t) with respect to t ... for example in the h seconds between time t = k and time t = k + h, the ball drops an extra s(k + h) - s(k) meters ... so the average velocity between times t = k and t = k + h is (s(k + h) - s(k)) / h, and the limit as h -> 0 of (s(k + h) - s(k)) / h is the velocity at time t = k and is the derivative of s(t) at t = k, and you can see that it has units meters per second because s(k+h) - s(k) is in meters and h is in seconds.\n\nSo distance is meters and velocity is meters per second, with v(t) = s'(t) the derivative of s with respect to t.\n\nNow acceleration is defined as the rate of change with repsect to time of velocity.  You will have a(t) = v'(t) = s''(t), by the same reasoning as before.  The velocity at time t = k is v(k) = s'(k) and the velocity at time t = k + h seconds is v(k + h) = s'(k + h).  The average acceleration between times t = k and t = k + h is then (v(k + h) - v(k)) / h in units of (meters per second) per second, or meters per second squared.  The instantaneous acceleration at time t = k is the limit of that as h -> 0 and is v'(t) = s''(t) at t = k.\n\nFalling in a vacuum at constant acceleration g = 9.8 meters per second squared means you have\n\nacceleration = constant g = 9.8 meters per second squared\n\nincreased downward velocity after falling t seconds = g t = (9.8 meters per second squared) time (t seconds) = 9.8 t meters per second.\n\ndistance fallen (if dropped with initial velocity zero) after falling t seconds is (1/2) g t^2 = (1/2) (9.8 meters per second squared) (t seconds)^2 = (1/2) g t^2 meters.\n\nThe rate of change of acceleration would be meters per second cubed.
934381
Basically you know the initial velocity that the baseball starts out with in both x and y.  Let V = 25m/s.  For A:  Vxo (initial velocity in the x direction) = 25*cos(60).  Vyo (initial velocity in the y direction) = 25*sin(60).  You know this similarly for ball B with a different angle.\nYou can use your kinematic equations\nd = Vo*t + (1/2)*a*t^2\nVf^2 = Vo^2 + 2*a*d\nVf = Vo + a*t\nd = (1/2)*(Vo+Vf)*t\nin both the x and y directions to determine whatever quantities you want.  \nUse the knowledge that at height yA, the velocity in the y direction is 0.  Similarly for ball B.  \n\nThat should get you started.  Good luck!
234860
See the fig in the web site. Instead of lifting the load through a vertical height 'h' we are moving the load through a longer distace 'l' along the plane.\nIn any machine work will be the same.\nWork is the product of force and distance.\nIf the load is moved vertically we would have done a work Wxh.\nBut in inclined plane we move the effort P through a distance l . Work done is Pxl\n But P x l = W x h \nSince l is greater than h, P is less than w.\nTherefore \nAn inclined plane reduces the effort force by increasing the distance through which the force is applied\n\nIn your choise the second and fourth answer is  wrong\nas  no  (ideal) machine can increase or reduce our work. In an inclined plane effort distance (length of the plane) is more than the height. Therefore the first one is correct.
72876
He means does the angle of launch (theta) affect the range (R), when you launch an object at some constant speed (u)?\nThis is a basic question in applied mathematics (mechanics).\n\nYes it does.\nWrite the equations of motion for the object.\n(Neglect wind resistance, which is secondary.)\nAssuming the launch surface is flat and you did not launch at an elevation, thus y(0)=0,then the components of velocity in x-,y-directions are:\n u_x = u*cos(theta)\n u_y = u*sin(theta)\n\nNoting the acceleration: a_x(t) = 0, a_y(t) = -g (always downwards), then the velocity components are:\n v_x(t) = u*cos(theta)\n v_y(t) = u*sin(theta) - g*t\n\nHence distance traveled is:\n s_x(t) = u*t*cos(theta)\n s_y(t) = u*t*sin(theta) - 1/2 g*t^2\n\nThe object lands at some time t=t' when s_y(t') = 0 again (not the t=0 solution, obviously). Thus:\n 2*u*t'*sin(theta) = g*t'^2\n 2*u*sin(theta)/g  = t'\n...\nand the range R is the s_x traveled at this time t=t' : \n R = s_x(t')\n   = u*t'*cos(theta)\n   = 2*u^2*sin(theta)*cos(theta)/g\n\nNow u,g are constant here, hence range R=R(theta) is purely a function of theta:\n R = (u^2/g) * 2*sin(theta)*cos(theta)\n R = (u^2/g) * sin(2*theta)          ; [a trigonometric equality that sin(2*theta) = 2*sin(theta)*cos(theta)]\n\nIn this case you can show (without even needing differentiation) that R achieves a maximum when:\n sin(2*theta) = 1\n i.e. theta = 45 deg\n which is what you might have suspected.\n\n============================================================\n\n[For the general solution of launching at an elevation, y(0) = y_0 , then you need to rewrite s_y(t), solve as above, using differentiation to show R(theta) achieves a max when R'(theta)=0 for some theta which does not have a neat solution].\n\n[If you want to be a smartass, add the second-order air resistance terms into s_x(t), s_y(t), so you get acceleration due to a viscous force:\n a_x(t) =      -nu * v_x(t)^2\n a_y(t) = -g + -nu * sgn(v_y)*v_y(t)^2\n   [Note we have to use sgn(y_t) to ensure the viscous forces acts against direction of motion. Basically: down when the projectile is rising, and up when it's falling.\nBut you can usually neglect the air resistance component of a_y, since |v_x|>>|v_y| for most if not all of the flight, unless u is small and launch angle theta is large.]\n\nIf you neglect the air resistance component of a_y(t), then you can still get an analytic solution, otherwise I think now you get two difference equations and you have to solve approximately e.g. by Runge-Kutta approximation.\n\nIf you want to be an utter smartass, add in the effect of wind with a vector (w_x,w_y), such that the air resistance force depends on the relative velocity of projectile to wind:\n( v_x(t)-w_x , v_y(t)-w_y ) \n\nAlso, we neglected terrain (mountains etc.), these degenerate into piecewise tests to make sure you always maintain positive altitude s_y(t).\n\nIf you want to get silly, do the ballistic projectile case, where g decreases as a function of altitude, as does nu, and over long distances you have to factor in the earth's curvature, and probably you want to write it in polar, rather than Cartesian, coordinates.\n g(s_y) = g_0  - k_g(s_y(t))   (linearized approximation)\n nu     = nu_0 * exp (-k_nu* (s_y(t)+Earth's radius)) (atmospheric density)\n [but you can linearize this too, for the lower atmosphere anyway: http://www.aerospaceweb.org/design/scripts/atmosphere/ ]\n\nAnd remember this was just for a projectile, where mass=constant. For a rocket, mass is decreasing (non-linearly) over time, and the dm/dt change in mass is proportional to the acceleration force. The direction of the acceleration is of course along the current velocity vector.\nYou may have design limitations imposed on the maximum dm/dt (e.g. the fuel pumps have a maximum throughput). And you could have two- or three-stage rockets.\n\nIsn't it fun how complicated you can make simple things?\n\nHere's a
53108
PSEUDO WORK DONE\nThis is because of the pseudo work done by the inclined surface or plane.\nExplanation:- Let theta be the angle of inclination of the inclined smooth plane and h be its height.Think off an object(body) of weight W=mg is sliding down on the plane. The another force that is acting on the body is force offered by the plane against the weight of the body, let this be R, this force is acting perpendicularly to the inclined plane.Further due to the interaction of forces W & R there exists a resultant force by which the body is sliding down,let it be P,this force is acting along the line of inclination.Now in view of lami`s theorem we have, P=W sin(theta)= mg sin(theta);whence theta is the angle of inclination and 0<or equal to theta <or equal to pi/2.Hence it can be noticed here that sin(theta)is <or equal to 1.\nTherefore P<W ie. mg ( weight of the body).This clearly implies that force less than the weight of the body is enough to slide up to the height h through the inclined plane.For instance when Theta=0 inclined plane will be horizontal,consequently P=0.And when theta = pi/2 ; P will be = mg, this implies that the body falls freely.As a typical example let theta= pi/6.01...,P is< 1/2 mg.Therefore the force -P=-1/2 mg orP=1/2 mg is enough to slide the body to the height h.Accordingly work done by the force while lifting the body to h through inclined plane = 1/2 mgh \nMoreover, as h is the height of the inclined plane,Potential Energy (gravitation potential)at h is = mgh,which is also equal to the work done while lifting the body to the height h, by any path.ie. P.E=mgh. (P.E stands for potential energy).Here true work done is =mgh; work done by P = 1/2 mgh \nie. mgh-1/2mgh=1/2 mgh,This shows that there is a shortage of 1/2 mgh of P.E, when work is done through inclined plane.How to account for this shortage. Actually, there is no bargain in the laws of nature.\nTo account for the above shortage please once again consider lami`s theorem.\nie. R=W cos(theta)= mg cos(theta);where R is the force exerted by the plane on the body, perpendicular to the inclined plane.When theta is pi/6.01, R=1/2 mgh (nearly)\nTherefore pseudo work done by the plane = 1/2mg (nearly)\nie. work done by p + work done by R = True work done\n     1/2mg + 1/2mgh = mgh\nNote-plane is considered as frictionless.\nI think this is enough to understand the phenomenon.
74061
the simple answer is if you were to throw a ball at the groud 15 feet away from you the angle created between the ground and the path of the ball when its traveling towards the ground is teh angle of incidence\n\nhope this helps
1010100
ok, so if we're assuming that it was a head on elastic collision, and all the energy was transferred then there are two possible situations: if the 25lb ball was sliding then the total energy transferred will be equal to the amount it had originally (since we assumed elastic collision) that is equal to 1/2mv^2...where mass would usually be measured using the metric system. \nif the ball was rolling then it has not only translational energy (1/2mv^2) but also rotational energy which is equal to 1/2Iw^2 where I = inertia of a ball...if it is a solid sphere the moment of inertia is equal to 2/5mr^2 (m = mass, r = radius) and w (omega) = rotational speed in radians per second...which is equal to velocity/radius. so the amount of energy transferred would be equal to 1/2Iw^2 + 1/2mv^2.
796826
Velocity is a vector quantity.\nSo,if two velocities happen to be parallel to each other(same or opposite direction ),they are added algebraically.i.e.mere sum or difference as the direction would be same as the two given vectors.\n\nab----------------->      cd<------\nThen resultant   ------------------------->\n  +                                                         \n.............................................<--------\n=........................--------------->\n\nWhat made u think that resultant can be in diff. direction\nEven try using the triangle law .It gives the same result.
943006
The balance of forces acting on you: m*a=f1-f2; m is your mass, f1 is the force pulling you down the slope, f2 is force of friction, both being parallel to the slope incline. Thus f1=m*g*sin(b), f2=m*g*cos(b)*k, while g=9.8m/s^2, b=31deg, a=1.16m/s^2, m*g being your weight. Mind! While f1 is a vector parallel to the slope, the force m*g*cos(b) is normal to the slope and should not be taken into account unless friction be present. \nNow m*a=m*g*(sin(b)-k*cos(b)), hence k=(sinb-a/g)/cosb=0.46
495603
Assuming that the energy of the ball remains constant (i.e. no air resistance), the speed will be the same. The VELOCITY, however, will be -35 m/s.\nSpeed and velocity are NOT the same!
878371
The incline is a simple machine that permits less force to be applied to move an object. The work associated with moving an object on an inclined surface, in the absence of friction, is the same as raising the object of weight (m*g) through a distance H.\n\nFor the inclined plane with an inclination angle of 30 degrees, moving the mass (m) over a distance L would raise the object a distance H given by\n\nH = L*sin(30) = 0.5*L\n\nThe work then is\n\nW = m*g*H = m*g*L/2\n\nFor the values given\n\nW = (1.0 kg)*(10 m/s^2)*(0.5 m)/2\n\nW = 10/4 N-m = 2.5 N-m\n\nThe same answer can be found by looking at the net forces on the object and multiplying by the distance traveled.\n\nFu = 10 N (the upslope force)\nFd = m*g*sin(30) = (1 kg)*(10 m/s^2)/2 = 5 N (downslope force)\n\nFn = Fu - Fd = 5 N (Net force)\n\nW = Fn*L = (5 N)*(0.5 m) = 2.5 N-m
1131554
Yes. \n\n(All of the following bars air resistance.)\n\nA trajectory would be the path that it follows. If an object has no initial velocity, it would fall straight down.\n\nBut say it did have an initial velocity. Let's say you threw a football. Does the football ever stop in mid-flight and fall straight down? No; it has an initial velocity, a sideways trajectory. If something has an initial velocity sideways, it's trajectory will look like an upside down parabola (ie y= - x^2) once gravity is applied.\n\nHope that helps.
890152
Steepness is change in y divided by change in t. If y is distance and t is time, then this ratio is speed.\nP.S. that if you're going a constant speed then your y vs t is a straight line with slope equal to the speed.
614422
Depends what information you already have. The simplest one I can think of is:\n\nVelocity = Acceleration x Time\n\n-Therefore-\n\nAcceleration = Velocity / Time\n\nBut you might have other variables involved here.