a little object from heigh h release freely and at the last moment of falling, came down 44/1 m.indicate the heigh h (g:9/8 m/s2 )   (the answer:  122/5)
201642
Given v_i = initial velocity = 0; v_f = final velocity = 44 m/s; a = 9.8 m/s^2:\n\ndv/dt = a = 9.8 m/s^2\ndv = 9.8 dt\nint(dv, v_0, v_f) = int(9.8 dt, 0, t)\nv_f - v_0 = 9.8t\n44 - 0 = 9.8t\nt = 44/9.8 = 440/98 = 220/49\n\nint(dv, v_0, v) = int(9.8dt, 0, t)\nds/dt = v = 9.8t\nds = 9.8tdt\nint(ds, 0, h) = int(9.8tdt, 0, 220/49)\nh = 4.9(220/49)^2/2 = (49*22*220)/(49*49*2) = (11*220)/49 = (2200 + 220)/49 = 2420/49 ~ 49.3878\n\nAlternatively, since accelration was constant, you could always get it more quickly with the formulas that were derived with integrals as above:\n\nv_f = v_0 + at\ns = v_0t + 1/2at^2\n\nEither way you should get 2420/49 m if the initial conditions are correct.  I don't know where 122/5 comes from although it seems to be approximately a half of this answer.
636091
First divide up the initial velocity into its horizontal and vertical components.\nV is the initial velocity of 40 m/s.\nTheta is the launch angle of 37 degrees above the horizon.\nV_x_i and V_y_i are the horizontal and vertical components of the initial velocity respectively.\n\nV_x_i = V * cos(theta) = 40 * cos (37) = 31.95 m/s\nV_y_i = V * sin(theta) = 40 * sin (37) = 24.07 m/s\n\nAlthough the vertical component of the object's velocity will change as the object experiences gravitational acceleration toward the ground, neglecting air resistance, the horizontal component of velocity can be treated as a constant.\n\nV_x = V_x_i = 31.95 m/s\nV_y = V_y_i  - g*t\nwhere t is the time after the object is launched and g is the gravitational acceleration experienced by the object (g = 9.81 m/s^2).\n\nIf the fence which the object hits is 128 meters away (in the +X direction), then the object (traveling at 31.95 m/s in the +X direction) will hit it at about 4.01 seconds after launch.\n\nBy taking this value (4.01 s) and plugging it in for t in the below shown equation to solve for the distance relative to the initial launch height the object has traveled,\nd = V_y_i * t + 1/2 (g)t^2\none can calculate that the object is at a height of about 17.65 meters above its original launch height.\n\nSo I calculate the impact height to be 17.65 meters.
56851
Final veloc. when he reaches the land:\nV=sqr(2*g*h)= 8.743 m/s\nKinetic Energy= Ec= 0.5 * m* V2 = 1872,780 \nThis turns into potential Energy Ep= 1872,780  = mgh\nHe stops in h=0.8m , hence: \nForce = m g = 1872,78 / 0,8 = 2340,795 Newtons\n\nHe has two knees, so F= 2340, 795 / 2 = 1170,488 Newtons in each knee.
351188
Look, task 1,3,4 are very alike. We shall use the same formulas there:\n1. let us find energy of the body at A, at B, and at C\nat A:   mV^2/2 + mgH = m(V^2/2 + gH)\nat B H=0, so the energy constitutes the kinetic energy mV(b)^2/2\nat C: mV(c)^2/2 +mgH(c) = m(V(c)^2/2 + gH(c))\nthe energy does not change (not dissipates), so energy in A equals the energy in C:\n\nV^2/2 + gH = V(c)^2/2 + gH(C), so \nV(c) = sq.root {V(a)^2 + 2g(H(a) - H(c))} = \n = sq.root{64 - 2x2x9.81} = 4.976 m/s\n2. The force of gravity for the mass 100 kg will be \nF = mg = 100kg x 9.81 m/s^2 = 981N\nAs the mass is moved along steep surface, some part of the gravity force will hinder the movement. We can fin the component of the gravity force along the surface using the triangular 20m and 5 m proportion, (because the triangles are similar), so\nFg(along) = 981N x 5/20 = 981N/4 = 245,25N\nWe have to apply the force 300N to move the body along the surface, hence we overcome the gravity component 245.25, and the rest is for friction.  \nSo the Ffriction = F - Fg(along) = 300N - 245.25N = 54.75N\n\n3. The formulars are as in problem 1. \nat the start   Es = Ekinetic + mgH\nat the highest (final) point of the bounced ball\nEf = Ekineticf + mgHf,\nHere we have to make some remarks (very important)\nIf the question would be "How high the ball can bounce",\nthen we could assume that the highest point would be when the ball bounces vertically up, and at its highest point it has zero velocity, no vertical one, no horizontal one. Than we can makethe energies equal (assuming no dissipation) and find Hf:\nmV^2/2 + mgH = mgHf\nso,  Hf = H + V^2/2g = 4(m) + 36(m^2/s^2)/2x9.81(m/s^2) = \n             = 4(m) + 1.835(m) = 5.835 (m)\nIn case the ball bounces not vertically, but along the steep line, at his highest point it will have zero velocity too, so the answer will apply to the case. \n\n4. The velocity in B will be found using the formula for energy at starting piont and at B:\nmV^2/2 + mgH = mV(b)^2/2,\n\nV(b) = sq.root{V^2 + 2gH} = sq.root{16 + 12x9.81} = 11.56 (m/s)
420146
time when she hits the ground\nt=sqr(2d/g)\nt=sqr(2*7/9.8)\nt=1.19 sec\n\nx=vt\nx=(12m/s)(1.19s)\nx=14.28 m
364091
both are the same if it is drop at the same height because the gravity speed is 10 ms-2 (9.81ms-2 more accurate)
91871
see if this helps.....\n\n\n\nAnalysis of a simple gravity pendulum\n \nFigure 2. Force diagram of a simple gravity pendulum.To begin, we shall make three assumptions about the simple pendulum\n\nThe rod/string/cable on which the bob is swinging is massless and always remains taut; \nThe bob is a point mass; \nMotion occurs in a 2 dimensional plane, i.e. pendulum does not swing in and out of the page. \nConsider Figure 2. The blue arrow is the gravitational force acting on the bob, violet arrows are that same force resolved into components parallel and perpendicular to the bob's instantaneous motion, the motion along the red axis, which is always perpendicular to the cable/rod. Newton's second law\n\n \nwhere F is the force acting on mass m, causing it to accelerate at a meters per second2. Because the bob is constrained to move on the green circular arc, there is no need to consider any force other than the one responsible for instantaneous acceleration parallel to the known path, the short violet arrow in our case\n\n \n \nLinear acceleration a along the red axis can be related to the change in angle θ by the arc length formula\n\n \n \n \nThis, however, is not the acceleration we seek because the gravitational force on the bob causes a decrease in angle θ. One accounts for that by placing a negative sign in front of a, thus:\n\n \n \n \nFigure 3. Trigonometry of a simple gravity pendulum.This is the differential equation which, when solved for θ(t), will yield the motion of the pendulum. It can also be obtained via the conservation of mechanical energy principle: any given object which fell a vertical distance h starting from rest, would have acquired kinetic energy equal to that which it lost to the fall. In other words, gravitational potential energy is converted into kinetic energy. Change in potential energy is given by\n\n \nchange in kinetic energy (body started from rest) is given by\n\n \nSince no energy is lost, those two must be equal\n\n \n \nUsing the arc length formula above, this equation can be re-written in favor of \n\n \nbut what is h? It is the vertical distance the pendulum fell. Consider Figure 3. If the pendulum starts its swing from some initial angle θ0, then y0, the vertical distance from the screw, is given by\n\n \nsimilarly, for y1, we have\n\n \nthen h is the difference of the two\n\n \nsubstituting this into the equation for  gives\n\n \nThis equation is known as the first integral of motion, it gives the velocity in terms of the location and includes an integration constant related to the initial displacement (θ0). We can differentiate, by applying the chain rule, with respect to time to get the acceleration\n\n \n \n \nsame as obtained through force analysis.\n\n[edit]\nSmall angle approximation\nThe problem with the equations developed in the previous section is that they are unintegrable. To shed some light on the behavior of the pendulum we shall make another approximation. Namely, we restrict the motion of the pendulum to a relatively small amplitude, that is, relatively small θ. How small? Small enough that the following approximation is true within some desirable tolerance\n\n \nif and only if\n\n \nSubstituting this approximation into (1) yields\n\n \nSolution to this equation is a well-known, and quite expected, oscillatory function\n\n \nwhere θ0 is the semi-amplitude of the oscillation, that is the maximum angle between the rod of the pendulum and the vertical.\n\nThe term  is a pulsation, which is equal to ,\n\nwhere T0 is the period of a complete oscillation (outward and return).\n\nSince ,\n\nthe period of a complete oscillation can be easily found, and we have obtained Huygens's law:\n\n \n \n[edit]\nArbitrary-amplitude period\nFor amplitudes beyond the small angle approximation, one can compute the exact period by inverting equation (2)\n\n \nand integrating over one complete cycle,\n\n \nor twice the half-cycle\n\n \nor 4 times the quarter-cycle\n\n \nwhich
500194
I think F is the answer. Taking acceleration due to gravity is constant at 9.81ms^-2, but this assumes that air resistance is neglected, and g is a constant.\n\nIf air resistance is NOT neglected, and acceleration due to gravity is not taken to be constant, H is the correct answer. The rest of the options are wrong.
525056
NO.  All objects, when free-falling, fall at a constant acceleration of 9.80 m/s².  The formula to calculate the force of gravity on an object is Fg = m*g, where m is the mass of an object and g is the acceleratin due to gravity.  As you can see from this formula, since g must stay constant, if the mass increases, the force due to gravity must also increase.  Therefore, no matter what the mass of an object is, it will still accelerate towards the ground at the same rate.
397799
Impulse = change in momentum\n\nHenri falls 12 meters. The time it takes him to fall is t.\n\n12 = (1/2)gt^2 = (0.5)(9.8)t^2\n\nt = SQRT [ 12/(0.5)(9.8)] = 1.56 seconds\n\nIn this time, he will reach a speed of gt or (9.8)(1.56) = 15.3 m/sec\n\nAssuming he weighs 100 kg we can now calculate his initial momentum which is mv or (100)(15.3) or 1520 m-kg/sec\n\nHis final momentum is zero so the impulse from the water is just 1520 m-kg/sec
37063
A quick guess without having reviewed your math, try adding g = 9.8 m/sec^2 to your answer and see if that is the answer that is needed.
394234
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)
391675
though instant the bird leaves the swing we assume that the centre of gravity does not change of the system. Therefore 52*2=153*x, where x represents the swing velocity= 0.6797m/s.\n\nnow the energy for the swing is going to be conserved.  Just a great kinetic energy with potential energy and get the height. mgh=1/2mv2\n\nCalculate h.  That is the answer.\n\nIt took me quite a long time to type all this....so I hope you will give me the best answer.
482440
As is usual for these sorts of problems, we'll neglect air resistance.\n\nFor part (b) note that at the top of its trajectory, the velocity of the ball is zero, and the kinetic energy it had when it was first thrown has been convered to gravitational potential energy:\n\n0.5*m*(29.4 m/s)^2 = m*g*h\n\nwhere m is the mass of the ball\ng is the acceleration of gravity (assumed constant over the distance the ball travels)\nh is the maximum height\n\nSimplifying:\n\n432.18 (m/s)^2 = g*h\n432.18 (m/s)^2 / 9.81 m/s^2 = h\n44.07 m = h\n\nFor part a, write the position of the ball as a function of time, (assuming that the ground surface is at y = 0):\n\ny(t) = v0*t + 0.5*a*t^2\ny(t) = 29.4 m/s * t - 0.5 * 9.81 m/s^2 * t^2\ny(t) = t*(29.4 m/s - 4.9 m/s^2 * t)\n\nWe want to know the times when y(t) =0 (i.e., when the ball is at ground level).  Obviously, one of these times is t = 0.  \n\nThe other zero of this equation is found from:\n\ny(t) = 0 = (29.4 m/s - 4.9 m/s^2 * t)\n\n(29.4 m/s)/(4.9 m/s^2) = t\n5.996 sec = t
532859
The answer involves a bit of algebra and is easy to lose track of where you are.\n\nConservation of energy says that\n\n0.5*m*v^2*sin(q)^2 + m*g*h = m*g*H\n\nwhere H is the height at the peak of the object's travel, q is the initial angle, h is the starting height, v is the magnitude of the velocity, and g is the gravitation acceleration. Also, we know that\n\n(1): H - h = 0.5*g*T1^2\n\nwhere T1 is the time is takes to reach the top of its travel. This expression was obtained based on the fall from the peak to the starting height h. Using this expression and the conservation of energy expression\n\n(2): T1 = v*sin(q)/g\n\nThe remaining distance h is covered in a time T2 given by\n\n(3): h = v*sin(q)*T2 + 0.5*g*T2^2\n\nGiven the range of travel (L) we know that\n\n(4): v*cos(q)*T = L\n\nand the relation among the individual times is\n\n(5): T = 2*T1 + T2\n\nT1 is counted twice for the upward and downward motion of the arc. That is the time to reach the top of its travel from its initial height and to return to that height..\n\nUsing expression (5) and replacing T2 in expression (3) we have\n\n(6): 0.5*g*(T^2 + 4*T1^2 - 4*T1*T) + v*sin(q)*(T - 2*T1) - h = 0\n\nFrom equation (2) we can substitute for T1\n\n(7): 0.5*g*(T^2 + 4*v^2*sin(q)^2/g^2 - 4*v*sin(q)/g * T) + v*sin(q)*(T -2*v*sin(q)/g) - h = 0\n\nSimplifying this becomes\n\n(8): 0.5*g*(T^2 - 4*v*sin(q)/g*T) + v*sin(q)*T - h = 0\n\nUsing Equation (4) to replace T\n\n(9): 0.5*g*(L^2/(v^2*cos(q)^2) - 4*sin(q)*L/(cos(q)*g)) + L*sin(q)/cos(q) - h = 0\n\nwhich, after rearranging, becomes\n\n(10): g*L^2/2 - (h*cos(q)^2 + L*sin(q)*cos(q))*v^2 = 0\n\nor\n\nv^2 = g*L^2/(2*(h*cos(q)^2 + L*sin(q)*cos(q)))\n\nGiven the values you mentioned above, v = 9.125 m/s.
473909
In the vertical direction earth is pulling the object downward with a force.\n\nAnd all objects are pulled down by the earth with an acceleration given by g = 9.8 m/ s^2.\n\nFirst we consider the vertical motion alone.\n\nIn the vertical direction \nthe initial velocity u = 0.\nthe acceleration  a = g. \nThe time t = t to be found.\nDisplacement in the vertical direction   s  =   180 m.\n\nWe use the equation s = ut + 05 a tt.\n\nUsing this we find t = 6 second.\n\nNow let us consider the horizontal motion.\n\nThe object is released from the plane which was moving with a speed of 20m/s.\n\nThere fore all objects which was in the plane will also have a speed of 20m/s in the horizontal direction.\n\nHence in the horizontal direction the initial speed is 20m/s.\n\nThe acceleration in this direction is zero since no force is acting in this direction.\n\nThe time t is the same as 6 second.\n\nDistance we have to find.\n\nWe use the equation s = ut + 05 a tt.\n\nSince a = 0 the equation reduces to s = ut. = 20 x 6 = 120m\n\nThe horizontal displacement is 120 m.
486427
answer is approxomately 42.476m/s\n\nonly one formula here \nS=ut + (at^2)/2 \ns- distance travelled = 40m\nu- initial valocity\na - acc due to gravity-9.8m/s^2\nt- time \nin the first case the initial velocity is 0m/s^2\nso\n40=(9.8*t^2)/2\nt=2.857 sec\nas both the balls have to touch the ground at same time  and the second ball being dropped( rather thrown would sound correct) after 2 secs, the later has only 0.857 sec left for traelling a distance of 40m. \nso \n40=u*0.857+(9.8*(0.857^2))/2\nsolving u=42.476m/s\nthe initial velocity for the second ball should be 42.476m/s and should be thrown vertically downwards.\nthat is pheta=0 in the formula given un the question.