Explain acceleration due to gravity.?
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One common unit of acceleration is g, one g being the acceleration caused by the gravity of Earth at sea level at 45° latitude (Paris), or about 9.81 m/s².\nThe acceleration due to gravity denoted g (also gee, g-force or g-load) is a non-SI unit of acceleration defined as exactly 9.80665 m/s², which is approximately equal to the acceleration due to gravity on the Earth's surface at sea level.\n\nThe symbol g is properly written in lowercase and italic, to distinguish it from the symbol G, the gravitational constant, which is always written in uppercase; and from g, the abbreviation for gram, which is not italicized.\n\nThis conventional value was established by the 3rd CGPM (1901, CR 70). The total acceleration is found by vector addition of the opposite of the actual acceleration (in the sense of rate of change of velocity) and a vector of 1 g downward for the ordinary gravity (or in space, the gravity there). For example, being accelerated upward with an acceleration of 1 g doubles the experienced gravity. Conversely, weightlessness means an acceleration of 1 g downward in an inertial reference frame.\n\nThe value of g defined above is an arbitrary midrange value on the Earth, approximately equal to the sea level acceleration of free fall at a geodetic latitude of about 45.5°; it is larger in magnitude than the average sea level acceleration on Earth, which is about 9.797 645 m/s². The standard acceleration of free fall is properly written as gn (sometimes g0) to distinguish it from the local value of g that varies with position.\n\nThe units of acceleration due to gravity, meters per second squared, are interchangeable with newtons per kilogram. The quantity, 9.806 65, stays the same. These alternate units may be more helpful when considering problems involving pressure due to gravity, or weight.
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Well, yes, the gavity is pulling you down, and you are not moving. So you are subjected to an acceleration and are not moving.\nMind you, one can argue that the *resulting* force is zero, as the ground is reacting with an equal and opposite force to your weight; if you take it that way, to have an unbalanced force should result in motion.
28151
Acceleration due to gravity (NOT change in acceleration), which could also be described as change in velocity.  That's assuming no air resistance.  If there is air resistance, they will eventually reach terminal velocity after which their velocity will be constant, but this is not implied by "free fall," and given that the question sounds like it's from a basic middle school or high school physics class I'm guessing you're supposed to ignore air resistance.
630060
Speed of.. gravity? :S
46446
it derives from equation x=0.5*a*t^2+v*t+x0 (I know equation in metric, maybe 0.5 isn't correct) where a is accelarion and v is start speed and x0 is starting place, a has been replaced by -32 which is gravity acceleration and v is assumed 0 and x0 by s.\n weight doesn't have any effect if there is no air pressure(usually you can think so) , it is famous that Galileo Galilei famous Italian scientist proved it from top of Pizza building with 2 different balls, they reach to the ground together.\nThe effect of air is a little complicated, it depends on the way you simulate it but usually it slow down the accelerate till the point that speed remains constant. Depends on the density of air and the shape of thing.\n there isn't any equation  which describes such a phenomena, you must solve the problem case by case.\nIf you like dynamic and static physics I recommend you The physics by Holliday, its a good book.
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Keep in mind these equations\n\nx(t) = x-initial + (v-initial)t + (1/2)(a-gravity)t^2\nv(t) = v-initial + (a-gravity)t\n\n(v-final)^2 = (v-initial)^2 + 2g(x-final - x-initial)\n\na-gravity = -9.8 m/s = -32 ft/s (i think..)\n\nTherefore\n\nx(t) = 0 + (64)t - (16)t^2\nx(t) = -16t^2 + 64t\n\n(v-final)^2 = (v-initial)^2 + 2g(0)\n(v-final)^2 = (v-initial)^2\n\nsince it is going in the opposite direction,\nv-final = -(v-initial)
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9.81 m/s^2 is the acceleration due to gravity on the Earth's surface, it is a constant.\nIf you were to plot the acceleration of an object as it undergoes a constant acceleration, the slope of the line would be zero.
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The acceleration due to gravity is independent of the mass of the object.\n\n The speed of any mass thrown up will decreases by an amount 9.8 m/s per second. \n\nIf the initial speed is u, then its kinetic energy is ½ m u u. Its potential energy is zero.\n\nThe total energy is½ m u u + zero = ½ m u u.\n\nAs it goes up its kinetic energy decreases whereas its potential energy increases keeping the total energy same.\n\nWhen it goes up through ‘h’ meter,  its total energy is [ ½ m u u + mgh].\n\n[ ½ m u u + mgh] = ½ m v v.\n\nm u u + 2mgh = m v v. \n\nCancelling m on both sides,\n\nu u + 2gh = v v\n\nThus we have the familiar equation vv = uu + 2gh.\n\nWhen the ball reaches 9 m its speed becomes zero.\n\nHence,  v v = 2 x 9.8 x 9\n\nSolving we get v = 13. 3 m/s . \n\nThe speed with which it is thrown up is 13. 3 m/s.\n\nWhen the height is 8 m, now having known the initial speed we use the same equation for solving for uu.\nvv = uu + 2gh\n\n13.3 x 13.3 = uu + 2 x 9.8 x 8 .\n\nSolving for u, we get  u u = 196 or u = 4.43 m/s.\n\nThe problem can be viewed in other way.\n\nWhen the ball is at its heighest point namely at 9m its potential energy is mgh. where h is 9m\n\nWhen it falls through one meter its kinetic enery increases and potetial energy decreases , keeping the total energy as the same.\n\nmg x 9 = ½  m vv + mg x 8.\n\nOr mg ( 9-8) = ½  mvv \n\ng x 1 = v  v/2\n\n2 g = v v. \n\nv v = 2 x 9.8 = 19.6 (m/s)^2. \n\nv = 4.43 m/s.\n\n Note that the mass gets cancelled and hence we need not know the mass of the ball.
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It seems the 16 m/s is not relative to this question. That is velocity in the horizontal direction. Gravity only effects in the verticle in this case. \n\ntime = sqrt(2 x height/ g) = sqrt(2 x 45 x 9.8) = 3.03 seconds\n\nvelocity = time x acceleration of gravity = 3.03 x 9.8 = 29.694m/s
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B.  The acceleration is towards the Earth and the force is constant.
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this is the amount of force.  Force equals mass times acceleration.  The mass is ur bodyweight in kilos and the acceleration is gravity 9.8m/s/s
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a speeding object slows down in time due to gravity not speed. that's why steins general theory was also trying to tell us. that gravity slows time.
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Extreme position! \nBecause acceleration is directly propotional to the displacement from the mean position and directed opposite to it.\n\na=-(ω^2)x
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Newton's second law is \nforce = mass x acceleration\n\nHow about this? Two men were walking on a hill when they see a bar of chocolate at the bottom. Both are eager to get it. The first man, a skinny little guy of 120 lbs, trips and begins rolling down the hill. The second man, a fatty of 250 lbs, trips at the same time and begins rolling down the hill. Because the fat man has more mass, he is able to produce more force to accelerate himself down the hill faster than the skinny man. The happy fat man reaches the bottom of the hill first, and eats the chocolate.\n\nIllustrations shouldn't be too hard.
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You don't need the mass, just multiply u*g, to calculate the deceleration. After that use this equation:\n2*a*d = Vo^2, where a=u*g, u=coefficient of friction, g= gravity acceleration.\nVo initial velocity, then\nVo= sqrt(2*a*d) maybe the velocity was greater than that value.\nI hope this is useful for you.
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gravitational potential energy\nm*g*h where m=the mass g= the acceleration of gravity which is(9.8m/s^2). h= the height.
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Force = mass x acceleration\n Mass = weight times gravity\nAcceleration is speed/time\n example if I’m playing football and I want to move the other player assuming we are both on the same planet (gravity is constant) and we are both moving at the same speed at each other I can do so if am heavier than him (increase wt (g) = increase mass) or I decease the time it takes to reach that player (constant speed and distance in less time = acceleration) or I can do both. This is how you increase force by increasing acceleration, mass, or both