What is the equation for the force of a falling object when it hits the ground from a known height?
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
1103152
Falling objects do not fall at a constant speed, they accelerate (get faster) the longer they fall. In a vacuum (no air resistance) two objects of the same mass (weight) experience the same acceleration. So if they're dropped from the same height, they will hit the ground at the same time. \n\nIn the atmosphere, air resistance will diminish the force of gravity, so two objects of different shape or density would accelerate at different rates because of different air resistance.
924813
(288^2)/(2*32.2) is the maximum height.\nat time 288/32.2 seconds\nit hits ground at time time   t+576/32.2, where t is obtained from\n45=288+.5*32.2*t^2
1120113
The miximum height reached by the ball is when the velocity of the ball is zero.  Since the initial velocity is 64ft/s, the distance above the ground equation becomes s = -16t^2 + 64t + 0\ns = -16t^2 + 64t .  Taking derivaties to find velocity equation.\nds/dt = -32t +64.  At maximum height ds/dt = 0.  \nThus, -32t + 64 = 0\n            32t = 64\n              t = 2\nSo, maximum height is reached at t = 2s.  So the maximum height reached by the ball:\ns = -16(2)^2 + 64*2\ns = -64 +128\ns = 64 feet.\nThe maximum height reached by the ball is 64 feet.
434934
I am not sure I understand your question, but here goes. If something is higher it has more potential energy than if it is lower. If you drop something from 10 stories high it will get damaged. If you drop something from only and inch it will probably be all right. It also matters how big the object is so PE (potential Energy)=m(mass)xh(height). Hope this helps.
955270
I won't do it for you, but here is what you need to do:\n\nFind the height at which the explosion occurs = h.\n\nFind the horizontal component of momentum at the instant before the explosion (which is actually the entire momentum, since there is no velocity in the vertical direction at that instant!) = Mtot x Vhoriz =m1 (Mtot is the entire shell).......(1)\n\nNow, the total momentum the instant after the explosion is:\n\n(M1 x 0) + (M2 x V2) .......(2)(since the large part of the shell stops travelling horizontally, and is already not travelling vertically) Where V2 is the horizontal velocity of the small part of the shell.\n\nSince momentum is conserved, these two quantities are equal, so:\n\nMtot x Vhoriz = M2 x V2, so you can calculate V2, the new horizontal velocity of the small part of the shell.\n\nNow, simply calculate the time taken for the shell to travel vertically to the ground using s = 1/2at^2, and from V2, you can calculate the horizontal distance travelleed by the small part of the shell...presto!
341507
You mean find "g" as if you didn't know the constant? Okay.\n\nPretty simply really. Have the height measured that you're going to drop it from to where it's going to land (this will be your "h"). Make it a high one (one meter is enough). Measure the mass of the object (that's the "w"). Then when you do the experiment, time it. Start it at the instant you let it drop and stop it once it touches where ever it lands. That's enough information to find "g." Just solve for "g" then that would be the answer.\n\nActually, for an experiment, you should solve for "g" as a formula before you start doing anything else. So the formula would be like this:\n\ng = (2h)/[(t^2)(w)]\n\n(t^2 is "t squared")\n\nSo just have that formula then do the experiment: measure height of the drop, the mass of the object, then time it when you drop it (you need good reflexes when you drop it, as it falls fast).\n\nPS Penguins!!!
434509
Here is what you want to know. Toss an object into the air. The point at which it changes direction is the point at which the gravitational force on it is exactly equal to the force you applied when you threw it. This is the maximum height. At that point only is it "weightless" immediately before and after it is under the effects of gravity.
835975
This question can be answered with both calculus and standard constant acceleration equations.  In constant accelaration, velocity equals acceleration multiplied by time (v = at). Taking the integral of this gives us popular phyics equation d = v0t + (1/2)at^2, or distance equals initial velocity multiplied by time plus half the acceleration multiplied by the time squared.  Using these equations we have:\n\nAt peak of the dive:\nv = v0 - at\n0 = 12 - 9.8t\n9.8t = 12\nt = 1.22 (time in the air at peak)\n\nHeight at peak:\nd = vot + (1/2)at^2\nd = 12(1.22) + (1/2)(-9.8)(1.22)^2\nd = 7.34684 feet\n7.34684+10 feet (height of board) = 17.34684 feet above water\nSo, the answer to A is 17.347 feet\n\nWhen she will hit water:\nd = (1/2)at^2\n17.34684 = (1/2)(9.8)(t^2)\nt^2 = 3.54017\nt = 1.8815 (falling) + 1.22 (to reach peak) = 3.1015 seconds\nSo B is 3.1 seconds\n\nShe was falling for 1.8815 seconds at 9.8 m/s, so using v = at we have:\nv = 9.8(1.8815)\nv = 18.4387 m/s\nSo C is 18.4387 meters per second.
230718
Answer:\n8 seconds\n\nProof:\nThe problem is really a question of finding what value(s) of x make y equal to zero in the given equation.\nWhen y (the height or distance) equals 0 then the ball is either leaving the ground or hitting the ground. The values of x (the time) that make the height equal to zero tell us when this is happening.\nThe given equation is an inverted parabola so there may be two points where y = 0.\n\n1) Solution 1: x = 0\ny = -16x^2 + 128x\ny = -16*(0^2) + 128*0\ny = -16*0 + 0\ny = 0 + 0\ny = 0\nBut x=0 refers to the time when the ball is first thrown. We want to find the time when the ball lands.\n\n1) Solution 2: x = 8\ny = -16x^2 + 128x\ny = -16(8^2) + 128*(8)\ny = -16*64 + 1,024\ny = -1024 + 1,024\ny = 0\nWhen x (the time) equals 8 seconds, y (the height) equals 0. So ball hits the ground after 8 seconds.\n\nFor your information, after 4 seconds the ball reaches its maximum height of 256 feet.
678477
9.8 x .020 will give you the gravitational force between the balloon and the earth.  The electrostatic force between the balloon and the ceiling could be different but must be at least 9.8 x .020 (otherwise the balloon would fall).  If the electrostatic force was larger than the gravitational force, there would be a normal force from the ceiling pushing down on the balloon (otherwise the balloon would try to calapse into the ceiling).
655249
I know the following two ways:\n\n1) Force on a body  is directly proportional  to the product of it's mass and acceleration, F=K*m * a(k=constant).The units are so chosen so that K=1.\n Your's is also correct but it is limited to free falling bodies, where a=g=9.8 m/s^2.\n2)Force is directly proportional to the rate of change of momentum.\nF=k*(change in momentum,p)/t=K*p/t. Again K=1 for same reason..\nBoth are related as p=m(v-u)/t=m*a.\nv=final velocity, u= initial velocity.\nhope it helps.
466570
Mass.\n\nF=ma\n\nF is the force\nm is the mass\na is acceleration
866141
You have to calculate p=massXspeed before the "collision".  The quantity remains constant, so if the mass increases, the speed must decreases by the same ratio for the p to stay the same.  (First, you have to convert the 8N weight to a mass using the 9,8N/kg gravity constant)
125809
The struggle here is that you provide too little information to recognize this problem as coherent.\n\nWe may assume that the acceleration of gravity is 9.8 m/s/s or 32 ft/sec/sec.\n\nThe velocity(V) of an object is equal to its initial velocity(v) plus its acceleration(a) times time(t). \nV = v + at or V = v - 32t (in feet per second)\n\nYour object will reach maximum height when the velocity is zero or when the initial velocity is equal to 32 ft/sec/sec * time. With your list of times, you'd just multiply by 32 to solve for the initial velocity, if that is what the problem is actually looking for. There are many other permutations of this data, but hopefully that is a start for you.\n\nGood luck.
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
833368
find the force of her weight. \nthat should be the answer