if a body is projected upwards with an intial velocity v0 prove v*v = 2gr*r/s+v0*v0-2gr?
467967
this is probably not a perfect answer but this site may give you some clues.\n\nhttp://www.gomath.com/Questions/question.php?question=4751\n\naltrnatively type 'prove v*v = 2gr*r/s+v0*v0-2gr' into a search engine and see what comes up.
449048
P1 x V1 / T1 = P2 x V2 / T2\n\nYou have all you need.  Calculate it.
338257
a) vcenter = (m1*v1+m2+m2)/(m1+m2)\n\nb) v1 rel = v1 - v center\nv2 rel = v2 - v center\n\nremember velocity is a vector so you must keep track of magnitude and direction
162313
Let V = speed (the magnitude of the initial velocity)\na = angle, t = time, \nVx = horizontal (x) component of velocity\nVy = vertical (y) component of velocity\nAx = acceleration in x direction\nAy = acceleration in y direction\nDx = net distance traveled in x direction\nDy = net distance traveled in y direction\n\nVx = Vcos(a)\nVy = Vsin(a)\n\n  Dx = 1/2 Ax*t^2 + Vx*t\nbut Ax = 0, so Dx = Vx*t, t = Dx/Vx\n\n  Dy = 1/2 Ay*t^2 + Vy*t\n(Ay = -g, the acceleration of gravity, negative for downward)\n -1/2*g*Dx^2/(Vcos(a))^2 + Vsin(a)*Dx/(Vcos(a)) = Dy\nMultiply by V^2, and collect terms with V^2 on the left:\n(Dx*tan(a)-Dy) V^2 = 1/2*g*(Dx)^2/(cos(a))^2\nV = sqrt(1/2*g*(Dx)^2/(cos(a))^2/(Dx*tan(a)-Dy))\nsubstitute a = 30 degrees, Dx = 6.05 m, Dy = 3.05-2.2 = 0.85 m, g = 9.8 m/s^2\nV = 9.51 m/s\nFor a = 60 degrees,\nV = 8.63 m/s
798759
1- yes, 2-yes, 3-no Vx=the same 24.5m/s, 4-yes, 5- no Dy=21, 6-yes Ax=0 throughout the whole parabola. 7- yes coz the only force (constant force) is gravity. Now (8) – its final Vy=26, and Vx=  still the same 24.5, thus Vtotal=sqrt(Vx*Vx+Vy*Vy)
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)
976194
5,25rad/s = 2πf\nf= 0,836Hz\nVt= (2πr*f)\nVt= 2π*0,0600m*0,836Hz= 0,315m/s\nAc= (V^2)/r\nAc= (0,315m/s)^2/0,0600m= 1,66m/s^2\n\nBoth points have the same frecuency, since they have the same angular velocity. So, for point 2 you already have a velocity of 5,25rad/s.\nThen, Ac= (V^2)/r →  (4π^2*r^2*f^2)/r→ (2πf)^2*r\nTherefore, Ac α r.\nIf r is half, then Ac is half (1,66m/s^2)/2= 0,83m/s^2\nthe same for the velocity.\n Vt= (2πr*f), because the radius is half and Vt α r\nVt= 2π*(r/2)*f = π*r*f= π*0,0600m*0,836Hz= 0,158 m/s\nOr just (0,315m/s)/2= 0,158m/s
303805
I don't see how components are necessary.  \n\n"r" and "v" are just shorthand for r(t) and v(t), assuming we are looking at position (r) and velocity (v) as time varying "vectors".  And v(t) is just the time derivative of r(t) (by DEFINITION), that is, v(t) is DEFINED as r'(t).\n\nSo there isn't any actual CONTENT in expanding them into components. We are just choosing different ways to symbolize the quantities involved and there isn't anything to prove.
844762
There are jthree  formulae cncerning velocity.These are v=u+at , V^2=u^2+2as ,and s=ut +(1/2) at^2 ,where u= initial velocity,v=final velocity ,a=accelaration.,s= distance travelled and t= time.Now how to decide which formula is to be used?in the problem,just see what they want you to find out and note down what is are already given in the problem, if they give you u,a and t and ask you to find out v,apply formula no.1.But if they gdonot give you time but give the distance or displacement and asks you to find v,donot hesitate to apply formula no,2 and so on.if you have any difficulty to understand.please do not hesitate to write to me through yahoo answersand i shall be glad to assist you
855600
u can also solve this by takin s= -100m\nn by the formula S= ut + 1/2 gt^2\ng = 9.8m/s^2. u = -6m/s .......4 the 2nd part\nand \n\nfind time of the 1st half by the formula \n=> v = u + at....[ where u = 6m/s, v = -6m/s(velocity in oposit direction)]\n\nALSO REMEMBER THAT WEN A TRAJECTORY AGAIN REACHES ITS STARTIN PT.  IT HAS THE SAME MAGNITUDE OF VELOCITY\n\n-ve sign is given to all dat is below the level or in the direction below the startin point\ncombine these 2................ n voila..\nwith time u can master this easy.... n logical method
680534
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
321671
Ah, this brings me back!\nThis is conservation of momentum problem. That means that the initial momentum must equal the final. You must break up the values to x and y directions.\n\nThe 900kg mass is traveling 18mps * cos(30) = 15.6mps in +x\n                            18mps * sin(30) = 8.98mps in -y\nThe 750kg mass is traveling 0mps in x\n                            23mps in +y\n\nWith these values take the initial momentum as\n\ninit mom in x = 900kg(15.6mps) + 0 = final mom = 1650kg*(v)\nwhere v is final speed in x = 8.5mps\n\ninit mom in y = 900kg(-8.98mps) + 750kg(23mps) \n\nthat equals the final mom in y = 1650kg*(v)\n\nso v=+5.6\n\nIf you know the speed in the x and y directions, convert to vector notation by using s^2=vx^6+vy^2 = 10.2mps\nand the angle is atn(vy/vx)\n\nHope this helps
754995
Nicole I think that to understand what is to be done here I suggest to assume that a coordinate system be attached to one of the objects. I-don’t-remember-who says, but << any reference system at rest or moving with linear constant speed has no privilege with others the like>>.\n(x) m2=4kg is reference system #2, where v2=0, and m1=2kg is moving with speed v=8+12=20 ms, impact by m1 on m2 M1=m1*v=40kg*m/s\n(y) m1=2kg is reference system #1, where v1=0, and m2=4kg is moving with speed v=8+12=20 ms, impact by m2 on m1 M2=m2*v=80kg*m/s, so R=40/80=0.5
751049
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
792153
‘impulse’ or momentum\n=m*(v2-v1)=0.42*(-17-4.1)=-8.862 kg*m/s
834761
Lets call the distance of the first part S1 and the total distance S2, and the corresponding times T1 and T2.  We know:\nS1 = 40\nS2 = 100\nT2 = 9.85\n\nIf V is the maximum velocity we know S1 = 0.5VT1 (const acc.)\nFor the second part we know S2-S1 = V(T2-T1) = VT2 - VT1\n\nVT1 = 2S1 so S2-S1 = VT2 - 2S1 or V = (S2+S1)/T2 = 140/9.85\n\nSo a) - V = 14.21 m/s\n\nWe know V^2 = 2AS1 where A is acceleration so A = V^2./2S1\nor A = 14.21^2 /(2x40)\n\nSo b) A = 2.525 m/s^2\n\nWe know V = AT1 so T1 = V/A = 14.21/2.525 = 5.62 s\n\nSo c) the two parts are 5.62s and 4.23s
958243
You need to know the initial speed.  Then the answer is:\n\n(v - u)/t\n\nwhere u = initial speed\n           v = 0 (final velocity)\n           t = 5 seconds\n\nThe answer will be negative because it is decelerating