Why free-fall acceleration can be regarded as a constant for objects falling within a few hundred meters?
406753
The average radius of the Earth is about 6378.1 kilometers, or abnout 6378100 meters.\nCompared to this, a few hundred meters doesnt make a significant difference.\n\nFor reference, the mass of the Earth is about 5.9742 E24 kilograms.\nG = 6.67 E-11\n\nThe gravitational acceleration on the surface of Earth (r = r_E = 6378100 m) is,\ng = G*M / r^2 = about 9.795 m/s^2\nIs the object is now 1 km above the surface of Earth (1000 meters higher), the gravitational acceleration is about 9.792 m/s^2, just .003 m/s^2 smaller, barely noticible.
733431
Because there is no absolute reference frame for motion, you can only define uniform motion with respect to a specified coordinate frame.\n\nFor example, a ball falling in a vacuum is not in uniform motion with respect to an earth-fixed reference frame; the ball accelerates.\n\nWhen the forces on an obejct sum to zero, that object is in uniform motion with respect to a reference frame that is non-accelerated. That means that if you compare gravity acceleration between the a frame fixed to the object, and the frame you are wondering about, if the gravity is the same, the frames are not accelerated with respect to each other.\n\nOthers have suggested that uniform motion requires absence of forces. This is not quite true - it requires absence of *net* forces. For example, a car traveling a constant velocity on a highway is not in a force-free environment: you can identify forces of air friction, propulsion from the wheels, weight, and support from the pavement. However it is considered to be in uniform motion with respect to a reference frame fixed to the earth, because all of those forces sum to zero, and because the apparent gravity in a frame fixed to the car is the same as in a frame fixed to the earth.
729735
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)
408862
the force between the object and the earth is dependant on mass, but the acceleration of the object towards the earth is 9.81 meters&#92;second. (ignoring air resistance)\n\nforce = mass x acceleration
796124
you will have to be in one point in the building to hear that the object will sound like a "musical scale going" down as it drops... but since objects on free fall accelerate, you would hear the 'musical scale going down" only at a very short time...\nso when you listen, listen carefully...
309904
There is a speed at which a falling object can't fall any faster, because the wind resistance holds it back.  It's called "terminal velocity".  It's different for a rock and a wad of paper the same size.  The terminal velocity for a raindrop is pretty low.
1054120
Assuming you meant to say "100 meters per second",\nyes, the object would.\n\nAfter the object is released it undergoes a fairly constant acceleration downward as its Kinetic Energy is converted into gravitational Potential Energy.  After the object reaches its highest point and begins to fall back down again (assuming no air resistance), all of the PE it gained on the way up is converted back into KE so when it gets back its original starting height, it has the same speed as when it left.\nThe law of conservation of energy at work.
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.
1163141
yes .. in vacuum, all items will fall at same rate... it is the rate at which the gravity acceleration attacts everything to the center of teh Earth. But you might have a queation, that if the acceleration  is same then why different things weigh different... \n\nweight =mass X acceleration of gravity (9.8meter/second^2) and mass is the one which is a measure of matter in a body makes the difference... so if you let a 5kg weight and a 5,000kg weight fall in a vaccum, they will fall at the same rate and will reach the ground at the same time.
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.
718152
weight equals mass times acceleration due to gravity
1183225
The instant an object is released as it is thrown upwards, it begins to accelerate downward.  This downward acceleration is caused by gravity (the force of weight, m*g, acting downward on the object).  Although the object may continue to travel upwards for a brief time, it is continually slowing in the upward direction as it accelerates towards the ground.\n\nObjects "fall" towards the ground (in the direction of the center of the Earth's mass) since that is the direction of the force of gravity is exerted on them.  An object cannot "fall" up since there is no net force acting in that direction.
223530
How about the adding of Gravitational Potential Energy (GPE) and Kinetic Energy (KE) as an object falls in the air to demonstrate the preservation of energy?\n\nBefore it falls all energy is GPE, as it falls GPE is converted to KE until just before it hits the floor then it is all KE.
826715
bullet drops 0.02 m due to gravity\n\nfree fall distance formula:\n\nd=v1xt + 0.5xgxt^2, v1=0 ( initial vert velocity is 0) and d=0.02 m\n                                   g=9.81\n\nt= (2x0.02/g)^0.5 = 0.0639 s\n\nv=25m/.0639= 391.5 m/s
87940
g ia actually 9.80m/s^2 for free falling bodies. And about the clocks pendulum, if you see it in a technical point of view, the answer would be NO; but it also depends on how the clock is dropped. The point is both the clock and the pendulum would freefall (9.80). Theoretically it's possible for the pendulum to swing but I can't explain it.
794040
This is really a calc question.\n\nAn item will fall at 32ft per sec per sec from a rest.\nh - the free fall = t\n\n2. As it is stated the axle is frictionless, therefore the cyinder and axle will move and there is no force between the two. However there is a force between the rope and or bucket and the cylinder.\n\ndouble check the 32ft/s in your book, there may be more to the equation.\n\nCarl
870312
a)For rigid body, center of mass accelerates in the direction of applied external forces. Just Newton's second law and therefore true.\nb)As in Newton's second law F=ma, force is dependent on acceleration and not directly to velocity only. So, there is explicit time dependence. hence, it really matters how fast the object gets to 10m/s or 20m/s velocity starting from rest. Therefore this statement is inconclusive with unknow time.