A round object can fall toward earth at what speed?
119178
Terminal Velocity.  It depends on the weight of the object because it is based on gravity.  If there is force applied to the object then it could travel faster for moments then gravity would slow it down.
386276
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m
236359
I will state this here, in yet another question: THERE IS NO SUCH THING AS CENTRFUGAL FORCE!! Look it up in a basic physics textbook like Giancoli or Serway. Or even do a google search. You have to understand what reference frame you're talking about. The fact that your body tends to want to move to the outside of the curve when your car goes around in a circle has nothing to do with anything other than the fact that your body wants to keep going along in a straight line (conservation of momentum), but the car gets in front of it at every turn (semi pun intended). What is actually happening is that the car is an accelerating reference frame.\n\nAs for the moon/earth question... imagine a man 6' tall throws a tennis ball horizonatally with respect to the earth's surface. The ball will go forward a certain distance (because of it's velocity tangential to the earth's surface) as it falls the ~6' to the ground (because gravity pulls it down). The faster he throws the ball (horizontally), the further it will go before it hits the ground. If he throws it fast enough then the amount that the ball travels forward and the amount that it drops towards the earth's surface can match the curvature of the earth, and the ball will orbit the planet (neglecting air resistance, as we physicists always do) and come around and bonk the thrower in the back of the head. The moon is just a much larger tennis ball, thrown by a much larger man.\n\nYour eqn. is correct, but it only describes the force acting upon the moon. It is not the equation of motion, which detirmines what actually happens to the moon. The equation of motion is:\n\nF = m * a  (F = force, m = mass of moon, a = acceleration of moon around earth)\n\nIn the case of the circular motion of the moon around the earth, "a" is defined by:\n\na = (v^2)/r (where v is the magnitude of the velocity of the moon and r is the center-to-center distance from the earth to the moon)\n\nPlugging one into the other you get:\n\nG * M * m/r^2 = m * (v^2)/r (where M is the earth's mass)\n--> G * M/r = v^2\n\nSo, as long as the tangential velocity, v = sqrt(G * M/r), the moon falls just enough towards the earth as it travels just enough tangentially second by second, that it continues to orbit around the earth. (Note: there is NO CENTRIFUGAL FORCE anywhere in any of these eqns.!!)
187092
Not unless the wind carries it because whilst the earth is spinning around rapidly, so is everything on it. If this were not the case, and a person jumped straight up into the air they would shoot of in a direction and this doesnt happen
41449
When an object is launched under the influence of only a uniform gravitational force field, it is said to follow a projectile motion. Under ideal condition, the path it traces is parabolic. There is catch here – if the initial velocity is equal or more than the escape velocity then the object will simply vanish into space i.e. leave earth!\n\nImagine if the launch angle from the horizontal is “theta” and initial velocity is “U” then you can easily compute vertical (Uv) and horizontal (Uh) components of it. The Uv will be equal to U x Sin(theta) and Uh will be equal to U x Cos(theta). Now by using Uv you can compute the max height it will reach and the total time it will take to go up and return to ground. With this round trip time in hand, you can easily compute the max horizontal distance it will cover before hitting the ground.\n\nAll these calculation assume ideal conditions like no wind resistance etc.
518923
My answer to the first part, by conservation of momentum, is that the speed equals 1.41m/s
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)
356458
All objects are pulled by earth with constant acceleration\n\nAcceleration is change in velocity per unit time = change in velocity / time\n\nIn one second the velocity changes by g unit.\nIn t seconds it velocity reduces by gt.\nIf u is the initial velocity and if in t seconds it is reduced to zero then u = gt.\n\nOr t = u/g.\n\nThe time to go up is u/g.\n\nThe height it has gone up is given by average speed x time. = u/2 x t = u/2 x u/g = uu/ 2g.\n\nWhen it falls down it starts with zero speed, but falls with the same acceleration g.\n\nWhen it reaches the ground let its speed is v and time of fall is t2.\n\nv= gt2.\n\nSince the distance is = average velocity x time = v/2 x t2 = gt2/2 x t2= gt2t2/2  \n\ngt1t2/2 = uu/ 2g. or t2t2 = uu/ gg  ot t2 = u/g .\n\nThus time of fall is equal to time of up.
309922
If we assume the Earth is a perfect spehere, the gravitational accleration would, indeed, be equal at any point along the surface of the sphere.\nWhen you take the rotation of the Earth into account, yes, at the equator, an object's "weight" is less than it would be at the poles.\n\nThe gravitational acceleration acts as the centripetal acceleration which keeps us all on on the Earth instead of flying out into space in a straight line.\nAt the poles, no centripetal acceleration is needed (neglecting Earth's revolution around the sun), but at the equation, the objects inertia acts to lessen the appearent effects of gravity and reduce its weight.  This is NOT due to the fictitious centrifuga force.
287995
Gravity won't let you. You have to go into orbit first, and 1 mph will not do that. You have reach escape velocity to escape earth's gravity, which is 11.2 km/s.
161915
approx 19 miles/second.\n\nAny force from centripetal acceleration you might feel would be more than compensated by the gravitational attractions from the Sun and Earth themselves.\n\nIt might be a measureable amount, but just barely.
292295
Consider two objects having masses 1kg and 5 kg are in a horizontal plane.\n\nWe apply a force of 10 newton to the 1 kg mass and 50 newton to the 5kg mass. \n\nThat is, we apply less force for less mass and greater force for greater mass.\n\nNow we find the acceleration of the two bodies.\n\nThe1 kg mass will have an acceleration of 10/1 = 10 m/ s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nSimilarly the 5 kg mass will have an acceleration of 50/5 = 10 m/s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nBoth have the same acceleration and both have the same speed at any time after start.\n\nNow returning to gravitational pull, let us place the 1kg mass in one pan of the balance and the 5kg mass in another pan. \n\nThe pan having 5kg mass will come down and the pan having 1kg mass will go up.\n\nThis is because the earth is pulling the 5 kg mass with a force of 50 newton and 1kg mass with a force of 10 newton.\n\nIf  we drop the two objects from the same place, as we have seen already the 1 kg mass will fall down with an acceleration of 10 m/ s^2 and the 5kg mass will also fall down with 10 m/ s^2. But the forces acting on them are different.\n\nThe Earth’s pull is such that it gives less force for less mass and greater force for greater mass.\n\nTo be more specific it gives a force of 10 newton for every kilogram of mass and hence the acceleration is for any mass is the same as approximately 10m/s^2.
50744
The object will follow a parabolic trajectory to the earth, as observed from outside the plane.  On the plane, it will appear to fall straight down in your (moving) frame of reference.  Air resistance will retard this motion, but it's magnitude depends on the drag coefficient of the object dropped and the medium in which it is falling.
330981
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.
495053
This is uniform motion, if it is going in a straight line at constant speed.\n\nIf there are no forces acting upon a body, and therefore no acceleration (remember F=ma) or change in direction, nor any change in angular momentum, the motion is uniform. If it were changing direction, accelerating or going round in circles then it is non uniform.
501881
acceleration due to gravity at a distance 4 times the radius is GM/(4R)^2 Where M is the mass of the earth . that is the answer is g/16.
663551
I think you have left off part of the question. The speed of a ball is a vector. When you hit the ball, the vector points upward and forward. Gravity pulls the vector downward. Small forces (air friction, etc.) retard the total speed, but assume those can be ignored.\n\nAt some point, the pull of gravity will pull the vector to point down instead of up. That's the maximum height. The horizontal component of the speed vector should not change (given our assumptions). It still goes forward at the initial velocity. However, The vertical component is instantaenously at zero. In a moment, the ball will start to accelerate vertically and the vector will continue to pull down below the horizontal line becuase of the force of gravity.\n\nIf this were not true, the ball would reach maximum height and then just drop from the sky (0 overall velocity) which you know it does not do.\n\nSo of your choices, probably equal to initial or ever so slightly les than initial would be the best.\n\nThe same thing happens if you drop, say, a ball from 10 feet and fire a bullet from a height of 10 feet. The both hit the ground at the same time.The fact that the bullet is speeding horizontally doesn't affect gravity pulling on it vertically. (yeah small effects like curvature of the Earth, wind resistance, etc. being ignored).