If an object is dropped from a helicopter, what happens when it reaches terminal velocity?
449788
Doesn't matter whether it's dropped from a helicopter, skyscraper, or the goodyear blimp.  Terminal velocity is the velocity at which the air resistance of a falling object is equal to the acceleration due to to gravity, and the object can no longer can gain speed.  It's not a constant, as you get closer to the ground air pressure will increase, so the air's resistance will increase, and the teminal velocity will drop. The object will actually slow down.
1157115
No, it falls back down at its terminal velocity(maybe), which is slower than the muzzle velocity because of drag.\n\nThe bullet leaves the gun with a given amount of kinetic energy: 1/2*m*v^2 where m is the mass of the bullet and v is its initial velocity.\n\nAir has non-zero viscosity, which is a resistance to shear forces.  When the bullet flies through the air, it drags air along with it.  This gives some of the bullet's energy surrounding air, reducing the bullet's velocity and kinitic energy.  Drag forces between fluids and solids occur in the direction opposite their relative motion.  (If the fluid is stationary and the bullet is moving, the drag is in the opposite direction of the bullet's motion.  If the fluid is moving and the object is stationary, the drag on the solid is in the direction of the fluid motion.)\n\nSimultaneously with drag, gravity is applying a force toward earth.  This too reduces the bullet's velocity and kinetic energy.  However, as the bullet flies higher into the air it gains an equal amount of potential energy from gravity.\n\nThe net result is that the bullet is always losing energy in flight because of drag.  This means that the amount of potential energy at the top is less than the initial kinetic energy, and it is thus impossible for the bullet to be going at the same speed it was shot when it hits the ground.\n\nThe fastest an object will fall under the influence of gravity is the terminal velocity.  At this velocity, the force of drag (which increases with the size of the object's profile, the viscosity and density of the fluid, and the relative velocity of the solid and the fluid) exactly balances the gravitational force.
850554
The answer is, there is an instant when a thrown object (upwards) reaches a velocity of zero, which is what the dropped object has the instant it is released. From these two instants in time, the response of each object is only to gravity.
384833
The terminal velocity of an object falling towards the ground\n\nVt= squareroot of 2mg/Cd pA\nVt is the terminal velocity, \nm is the mass of the falling object, \ng is gravitational acceleration, \nCd is the drag coefficient, \nρ is the density of the fluid the object is falling through, and \nA is the object's cross-sectional area
1161161
The law of conservation of momentum applies in all cases.\nFor a given system, the initial momentum must equal the final momentum.\n\nIn your example, we can define the "System" to be the combination fighter plane + bullet.  If we call this initial momentum P, then at any time after the initial conditions, the sum of the momentums of all parts of our system must add up to P.\nWhen the bullet is fired foward, the fighter plan slows down to compensate and conserve momentum.\n\nMomentum (P) = mass (m) * velocity (v)\nP = mv\n\nThe bullet's mass is very tiny compared to that of the fighter plane, so even a large increase in the bullet's velocity would only mean a small decrease in the resulting velocity of the plane.\n\nIf you call the bullet's mass, m_1 (with final velocity v_2) and the plane's mass m_2 (with final velocity V_2),\nwe know that,\nP = m_1*v_1 + m_2*v_2\nif P_1 goes up, then P_2 must go down by just that amount.\n(delta P_1) = - (delta P_2)\nm_1 * (delta v_1) = -m_1 * (delta v_2)\nsince we assume the mass stays constant,\n(delta v_2) = - (m_1 * delta v_1) / m_2
444669
There is a dent, although it is quite small and also short-lived. The air pressure just in front of the falling drop pushes the surface down just before the drop hits.\n\nBuh-bye now.
368139
If the object has onboard power, then the speed is not really relevant to whether it will leave the atmosphere, just when. If you just fire an object upwards and it has no self-propulsion, however, then you need to calculate the change in gravitational potential energy (find out how far the atmosphere extends) and add the energy lost to air friction (this depends on velocity). Equate this to the object's initial kinetic energy and solve for initial velocity v. Your initial velocity will have to be larger than this.
585763
When the lift descending,the acceleration is increasing.Relatively,body and the lift is moving down with a gradually increase speed.The acceleration will increase until its value is the same as the value of the acceleration due to gravity.\n\nLet R = normal reaction of the floor of the lift on the body, \n   mg = the weight of the body inside the lift\n    g = acceleration due to gravity\n      = 9.81 meter per second square (unit SI)\n    F = resultant force of the whole system\nusing this equation:\n\n     F = ma     \n     F = mg - R  (because the lift is moving downwards,\n                  resultant force is moving downwards too.)\nmg - R = m(a)    (a = g,because the acceleration of the lift\n                 is equals to the acceleration due to the\n                 gravity)\nmg - R = mg\n     R = 0\n\nAs R = 0,the body will not experience any force exerted by the floor of the lift,the body experiences weightlessness.\n\nIn short, weightlessness happens only when a body is in free fall motion, or fall freely under the gravity, or a more accurate explaination, weightlessness happens when a body falls freely with an acceleration which is equals to the acceleration due to gravity.\n\nHope my answer helps you.
1006314
No.  When you departed the aircraft you body would be travelling at the same speed as the aircraft was travelling.    Wind resistance would slow your body down very quickly but you would still land well ahead of the point in which you jumped out of the plane.
625418
As a skydiver falls (caused by gravity), he encounters the force of air resistance. The amount of air resistance is dependent upon two variables:\n\n1.  the speed of the skydiver\nAs a skydiver falls, he accelerates downwards, gaining speed with each second. The increase in speed is accompanied by an increase in air resistance. This force of air resistance counters the force of gravity. As the skydiver falls faster and faster, the amount of air resistance increases more and more until it approaches the magnitude of the force of gravity. Once the force of air resistance is as large as the force of gravity, a balance of forces is attained and the skydiver no longer accelerates. The skydiver is said to have reached a terminal velocity.\n\n2.  the cross-sectional area of the skydiver. \nA skydiver in the spread eagle position encounters more air resistance than a skydiver who assumes the tuck position or who falls feet (or head) first. The greater cross-sectional area of a skydiver in the spread eagle position leads to a greater air resistance and a tendency to reach a slower terminal velocity. The importance of cross-sectional area to skydiving is also demonstrated by the use of a parachute. An open parachute increases the cross-sectional area of the falling skydiver and thus increases the amount of air resistance which he encounters (as observed in the animation below). Once the parachute is opened, the air resistance overwhelms the downward force of gravity. The net force and the acceleration on the falling skydiver is upward. An upward net force on a downward falling object would cause that the object to slow down. The skydiver thus slows down. As the speed decreases, the amount of air resistance also decreases until once more the skydiver reaches a terminal velocity.
468173
it depends on how much the ground "gives". when it falls from a known height, it has potential enrgy=mgh, which translates to kinetic energy 1/2mv^2, so impact velocity is v = sqrt(2gh). final velocity (assuming inelastic collision) is 0. if you know how long it takes to deaccelerte to 0, the average force is f=ma=m*v/t = m*sqrt(2gh)/t. if you know how far it travels before stopping, v^2 = 2ax, a=v^2/2x, f=ma=mgh/x\n\n\ni realize an easier way to do this is to say that the work done by the ground is the same as the force on the object times the distance traveled through the ground. since \nWork = change in Energy\nEnergy = mgh\ndistance into ground = x,\nW = Fx\nW= E = mgh\nmgh= Fx\nF = mgh/x
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.
365737
Assuming you fire parallel to the ground, then the bullet will be at its maximum horizontal velocity when it leaves the barrel, and will be at minimum vertical velocity.  The bullet cannot increase in horizontal speed once it has left the barrel, as there is nothing to make it accelerate.\n\nDue to air resistance, the horizontal velocity will decrease as the bullet travels further and further.  The vertical velocity will increase, as the gravity of the Earth "attracts" the bullet.
985453
When it's falling and reaches a velocity where the force from drag is equal to the force from gravity. Simply put, it doesn't speed up any more.
433903
The ball will reach a maximum height, then the velocity of gravity (9.8 meters/second) would cause it to travel downward until another force acts upon it (according to Newton's laws of motion). However, this does not account for other forces, such as air resistance. The rate at which the ball falls and the air resistance acting upon it will depend on the properties of the ball (surface area, weight, etc.). If the ball is inside a vacuum, it is said to be in free fall (as the only force acting upon it is gravity) and there is no other force acting upon it. If the ball is allowed to fall long enough, it will reach a point of terminal velocity, where it cannot fall any faster (the force of air resistance will equal the weight of the object)
1046911
it will fall straight down , but appear to fall at an angle if observed from the plane
751261
It will accelerate rapidly to the surface of the water, and then continue to float upwards through the air, but with less acceleration.  This is all due to the buoyant force.  When an object is immersed in a fluid, the fluid exerts an upwards force equal to the weight of the displaced fluid.  Of course, gravity still acts downwards on the object.  But if the object is placed such that it displaces a volume of fluid with a greater weight than itself, as when the object is hollow or is less dense than the fluid, the net force is upwards.  Helium is less dense than air, and is much less dense than water.  Since water is more dense than air, the displaced volume of water weighs more than the displaced volume of air, so the buoyant force of the water is greater than that of the air.  But it is always a net upwards force, so the balloon is always accelerating, until it reaches its terminal velocity.  Terminal velocity is usually encountered with falling objects, but would also occur in an object accelerating upwards.