Neglecting Air Resistance, the rubber ball reaches the ground,\n\nA: at the same time as the metal sphere and at the same distance from the building\n\nB: at the same time as the metal sphere and at a smaller distance from the building\n\nC: after the metal sphere and at a smaller distance from the building\n\nD: after the metal sphere and at the same distance from the building\n\nPlease help me work out its answer.\nThanks
124567
The answer has to be A.\nThe time will be the same because Gravitational force does not depend on the mass of the object, since they start with same velocities and accelerate with the same rate they must hit the ground at the same time.\n\nAnd it obviously has to be the same distance because if it wasn't then they would never hit the ground at the same time in the first place. That would happen because if 2 bodies start with the same speed and have the same acceleration but different displacements then their time has to be different.
160441
I'm not sure what 500 above the horizontal means - radians maybe?  Or is there something missing, like a launch angle, and that you're 500 ft above ground when the ball is thrown?\n\nIf I'm picturing the scenario right, you're basically shooting a ball up at an angle and finding all of the specifics - max height, velocity at impact, etc.  All of those can be found using some of the following equations:\n\nvf^2 = vi^2 + 2ad\nvf = vi + at\nd = vt\n\nwhere\nvf = final velocity\nvi = initial velocity\na = acceleration due to gravity (9.8 m/s^2 or 32.2 ft/s^2)\nt = time\nd = distance\n\nHorizontal and vertical velocity are independent of each other.  The vertical velocity changes due to gravity, and the horizontal velocity will remain constant throughout the ball's flight (assuming no air resistance is in play).  So to find the max height, we'd need the vertical velocity component of the 110 ft/s.  That comes out to 110 sin x (where x is that launch angle I asked for earlier).  So this will be your vi in your first equation.  vf = 0 at the apex (the ball has stopped and is about to reverse direction), and since you are decelerating, use -32.2 ft/s^2 (I know you specified 9.8 m/s^2, but the units don't match the 110 ft/s - you can always convert one or the other if you like).  Solving that first equation will give you the distance it traveled.  If there is an additional height involved (remember if I asked if you were 500 ft above ground?) then your total height from ground level would be your answer from the equation plus 500.\n\nThe time to the apex can be found with the second equation.  Again, using vi = 110 sin x and vf = 0, you can solve for the time (and again, keep your units consistent or else the answer doesn't make sense).\n\nLet's jump to the last question about the velocity of the ball upon impact.  If it falls to the same height that it was thrown at, then the velocity on impact will be the same as the launch velocity - 110 ft/s.  But if there is that additional 500 ft height, we can use that first equation to find the vertical component of the impact velocity.  If we were to picture that part of the equation as starting from the apex, vi = 0.  The distance d will be the distance found in part (i) plus 500 ft.  So you can solve for the final velocity there - but remember that this only solves for that vertical component of the total velocity.  You'd need to find that total via the Pythagorean theorem - picture a triangle with the vertical velocity as one leg, and 100 cos x (the horizontal component of the original launch velocity) as the other leg.  Square both, add, and take the square root to find the total velocity on impact.\n\nOn to part (iii) - now that you have the final velocity (just use the vertical component as calculated above), you can use the second equation to find the time it took to fall from the apex to the ground.  Add this to the answer in part (ii) and you have the total time in flight.\n\nDown range, I'm assuming, is the horizontal distance the ball traveled.  Using the third equation, you can find that.  The velocity in this case is the horizontal component of the launch velocity - 100 cos x.  Take this times the time from part (iii) and you have the range.\n\nHope this helps - figure out what that launch angle is and how that 500 plays in to the equation and you should be on your way!
630795
First of all, if you can't phrase it properly in English, you'll never be able to get an answer.\nSecond, if the two balls (3x and x) collide and 3x has no further displacement, then all the energy of both balls must be accelerating x.  So you have xv☺ + 3xv☻ = 2xv☻ [I'm using ☺ and ☻ to demonstrate opposite directions.]  This follows conservation of momentum.
623895
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.
348740
(2) You need to consider conservation of energy. The energy of the ball at point a is\n\nm*g*Ha + 0.5*m*Va^2\n\nwhich, neglecting losses, is completely converted to kinetic energy at point b. Equating them gives\n\n0.5*m*Vb^2 = m*g*Ha + 0.5*m*Va^2\n\nand solving for Vb\n\nVb = sqrt(2*g*Ha + Va^2)\n\n(3) Same considerations, however when the ball reaches its maximum height (Hm) all of the energy is potential.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hm\n\nSolving for Hm\n\nHm = Ha + 0.5*Va^2/g\n\n(4) If the block is moving at a constant speed up the ramp due to the force F then summing the forces gives\n\nF - m*g*sin(q) - Fr = 0\n\nwhere q is the angle of the ramp, m is the mass of the block, g is the gravitational acceleration, and Fr is the friction force.\n\nSolving for Fr\n\nFr = F - m*g*sin(q)\n\n(5) Again, conservation of energy, but this time at point c the ball has both kinetic and potential energy.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hc + 0.5*m*Vc^2\n\nSolcing for Vc\n\nVc = sqrt(2*g*(Ha - Hc) + Va^2)
383339
Equation A is correct since momentum is a vector quantity it has direction.\n\nkinetic energy in equation B is scalar, so you should not subtract.\n\nI got the quadratic equation:\n\n75.83*V1^2-198.3*V1-904.167=0\n\nWhich gives two roots\nV1=5, or -2.384\nthe negative is the one I chose since the collision must be elastic and there is only one direction of motion (either positive or negative right or left).\n\nGiven this, I computed\nV1=-2.384\nV2=5.6\n\nj
184132
HAHAHAHA You all forgot about our little friend called drag! Inside or out the ball has a drag force acting on it by the surrounding air. Like stated before the air in the car is moving with you, and the car, and the ball all the same and the drag force on the ball at this point is negligible. This is not the effect of a vacume, a vacume is the absence of air. is it possible for there to be no air in the car? NO, Sooo.. outside on a bike or where ever, you are traveling at a speed where the balls light weight will be significantly affected by this drag force and it will cause it to slow down as you throw it up and while you keep riding on un affected by the static(motionless) air, your ball only has one force acting on it, air drag causing it to slow behind you.   test this by throwing a variety of same shaped objects while riding and you will see that the one with the most density will keep up the best. try a marble V.S. a ping pong ball.  then do it in the car
384368
Bob was travelling at 9.5 meters/sec when he started his jump. To gain maximum range he needed to jump at a perfect 45 degree angle. If he had done so, his initial velocity vertically would have been 9.5 /sqrt(2) = 6.72 meters/sec. His maximum height would occur at time t where 6.72-gt = 0.\n9.78t = 6.72\nt = 6.72/9.78 = 0.687 secs\n\nAfter that he would fall for another 0.687 secs till he hit the ground for a total of 2(0.697) = 1.37 secs\n\nIn this time, he would have traveled (6.72)(1.37) meters horizontally or 9.23 meters.\n\nSo his jump of 8.90 meters was 0.33 meters short of perfection.\n\nNote that air resistance (which is pretty important in reality) would have the effect of making the maximum achievable jump shorter than 9.23 meters. So his jump was better than it appeared..
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Its potential energy was 78.4 m joules before falling, where m is the mass of the ball.\n\nAfter collision when it reaches the height 5m its potential energy is 49 m joule only.\n\nIt has lost 8m joule of energy due to inelastic collision.
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use conservation of energy:\n\nThe 810 N force did work for 2 m\nthe result is potential energy gained in height working against gravity including the 2m that the 810N force acted on the ball\n\n810*2=7*9.8*(h+2)\nh=1620/(7*9.8)-2\n=21.6m\n\nj
131126
the first after collision moves in an opposite direction and the second ball moves towards the direction of collision\nfirst ball moves with a velocity of  16.66cm/s\nsecond ball with 66.66cm/s
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First Question:\n\na) Solving for the height of the 1st brick\n\nh=1.00m (sin 30.0)=0.50 m\n\nSolving for the velocity of the 1st brick at the bottom\n\nKE at the bottom = PE at the top\n1/2 v^2 = (weight) h\n1/2 v^2 = 24.3 (0.50)\nv = 4.93 m/s\n\nb) momentum before = momentum after; momentum is just m*v\n\n(24.3/9.8)(4.93)+(35.7/9.8)(0)\n=[(24.3/9.8)+(35.7/9.8)] [v]\n\nv = 2.0 m/s\n\nc) F = ma\n-5.0 = [(24.3/9.8)+(35.7/9.8)] a; F is opposite to motion\na = -0.817 m/s^2; deceleration\n\na =(vf-vi)/t; vf=final velocity; vi=initial velocity\n-0.817 = (0-2.0)/t\nt = 2.45 seconds\n\nd) 2ad = vf^2 - vi^2\n2(-0.817)s = 0^2 - 2^2\nd = 2.45 meters\n\nSecond Question\n\na)Make West East axis as the x-axis; N S axis as y-axis.\n\nx momentum before = x momentum after\n\n880(0)+1585(12) = (880+1585)(vx)\n\nvx = 7.72 m/s; x component of final velocity\n\ny momentum before = y momentum after\n\n880(-16)+1585(0) = (880+1585)(vy); compact car is headed in the negative y-axis.\n\nvy = -5.71 m/s\n\nv^2 = vx^2 + vy^2\n\nv^2 = 7.72^2 + (-5.71)^2\n\nv = 9.6 m/s answer\n\nangle = Arctan 5.71/7.72\n\nangle = 36.5 degress south of east\n\nb) F = coef of friction (weight of the cars)\n\nF = 0.58 (880+1585)(9.8)\n\nF = 14011 N\n\nF = ma\n-14011 = (880+1585) a; - because opposite motion\n\na = -5.68 m/s^2; decelerating (-)\n\n2ad = vf^2 - vi^2\n\n2(-5.68)d = 0^2 - 9.6^2\n\nd = 8.11 meter from the poin of impact
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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)
229194
It's conservation of energy.\n\nThing is that the ball loses energy by colliding on the ground, the ground absorbs some of the energy.  Then you also have gravity and friction in air... they all suck energy out of the bounce.
428320
the vertical component of the velocity applied to the ball is given by\n\nsin 35 = velocity/24.5\n\nvertical velocity is 14.05 m/s\n\nV^2 = u^2 +2gs\n\nat the maximum height, v = 0 so\n\n0 = (14.05)^2 + 2(-9.8)s\n0 = 197.48 + -19.6s\n19.6s = 197.48\ns = 10.07 metres\n\n\nmaximum height the ball reaches is 10.07 metres...
29731
a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
368277
the potential energy of the ball:\nPE = mgh = (.6 kg)(g)(.7m)\nwhen the ball collides with the block all the Potential Energy has been converted to Kinetic Energy so,\nPE = KE = (.6 kg)(g)(.7m) = (1/2)(m)(v^2)        (v^2 is v squared)\nall that kinetic energy is transfered to the block so\nKE(block) = 4.12 Newtons\n\nin one dimension the solution is that all momentum and kinetic energy is transfered to the block such that,\n(a) ball speed after collision = 0 m/s\nand \n(b) block speed after collision\n(1/2)(m)(v^2) = 4.12 Newtons\n(1/2)(2.5 kg)(v^2) = 4.12 Newtons\nv = 1.8 m/s