If something was dropped from 30 feet above the ground.  About how fast will it be going once it hits the ground?\n\n(In Miles Per Hour)
198948
This problem can be solved using v^2=2ax, where v is the velocity, a is the acceleration due to gravity, and x is the distance (this is a simplified version of a more complex equation, but this equation will work for this scenario).\n\na=32.15 ft/ s^2\nx= 30 ft\n\nSo, v^2=2(32.15ft/s^2)(30ft)\n\nOnce you solve and convert to miles per hour, you get approximately 29.9 mph.
664264
Let Time of fall be  t then\n 1/2 g t^2 = 25\nt= sqrt(2*25/g)\n\nIf original velocity was vh then distance covered \nvh*t=22\ntaking g=9.8\nvh =22/t = 22/sqrt(50/g)=22/2.258=9.73 m/s
49670
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.
466597
Well, how fast was it pushed?  Is it going uphill< downhill, on a flat surface, on a bumpy road, hmm, too many things to consider!
495208
First of all I assume the height of \nbasket above ground is 2+3.05=5.05\nLet v be the required velocity.\nvertical component=vsin40\n=v*.766\nfrom conservation of energy\nheight gained\n=[v*.766]^2/2g=3.05\nv^2=3.05*2*9.81/.766^2\n=101.986\nv=10.09m/sec
484875
So I'm assuming there is no static or kinetic friction in this problem.\n\nWell, since the cart experiences a force in the horizontal direction, then: \nF = ma\na = F/m\na = 11N/26kg = 11/26 m/s^2\n\nHow far will it move in 2 seconds starting from rest?  What do we know?\n\na = 11/26 m/s^2\nt = 2 secs\nVelocity initial = 0 m/s (since it starts from rest)\nSo what is d (for distance although I know some schools use X)?\n\nd = V(initial)t + 1/2at^2\nd = 0t + (1/2)(11/26)(2)^2\nd = 11/52*4\nd = 11/13 m\n\n30 N child in the cart before he begins to push it?\n\nNewtons is just another way of saying weight but in order to find the acceleration we need mass and not weight.\nWeight (in N) = mg\nm = weight/g\nm = 30/10\nm = 3kg  <-------- this is the mass of the child\n\nRepeat as above with little modification.\na = F/m\na = 11N/(26kg + 3kg) = 11/29 m/s^2\n\nd = V(initial)t + 1/2at^2\nd = 0t + (1/2)(11/29)(2)^2\nd = 11/58*4\nd = 44/58 m\n\nGood luck!
462106
(1000kg is not a ton, it is called a tonne or metric ton)\n\nThe object of 1000kg will have a force of acceleration on it of about 9.8m/s² during its fall of 5m.\n\nTo find the kinetic energy Ek that it will have at the end of its fall you just need to find its gravitational potential energy at the beginning of the fall.  By the end of the fall all the gravitational potential energy will have turned into kinetic energy.\n\nUg=(m)(g)(h)\n\nUg is grav. potential energy, m is mass, g is acceleration due to gravity, h is height\n\nUg=(1000)(9.8)(5)\nUg=49,000 Joules\n\nSo our Ek is also 49,000 Joules.\n\nThe object will land on the earth with 49,000 Joules of kinetic energy.\n\nNear the earth's surface, the object will have a force towards the earth of \nf=ma\nf=(1000kg)(9.8m/s²)\nf=98,000N\n\nAt the time of impact the composition of the earth will determine how much force the 1 tonne mass imparts on the earth.  If it lands somewhere soft where it can dig in far its kinetic energy will be eaten more slowly because it is pushing on the earth with a smaller force, if it lands somewhere very hard then the Ek will be eaten up very quickly as it will decelerate very quickly by imparting a large force on the earth.
421077
Wait, so you are saying he is moving 60mph, and he throws 60mph in the direction he was coming from?  If you sum the velocities you get zero; the ball would drop straight down (from your point of view).  To him, it would appear to be moving at 60mph.\n\nIf he simply dropped the ball, it would be moving at 60mph alongside him.  To him, it would look like it were falling straight down.  You can test this; toss a ball straight up in a moving vehicle - as long as you do not turn a corner or change speed, the ball will move straight up and come back down.\n\nObviously wind resistance at 60mph is significant, but since you said it was in a vacuum, there is no wind.  If you tried tossing a ball up in a convertible with the roof open, it would decelerate a bit.
350015
Acceleration on any free falling object is always equal to the force of gravity, which is 9.8m/s^2.\n\nSince acceleration is a vector quantity, it can be denoted as positive or negative. In this case, since the object is moving downwar and not up, acceleration would be negative and so equal to -9.8m/s^2.\n\nGood luck.
505188
please make your question more understandable\n\nif it decreases from 30mph to 15mph over 75 metres then at this deceleration its logical to say it would take another 75metres to stop\nbut something tells me thats not the answer ! if you want me to do your homework at least have the decency to type in the full question.
16821
How high up is the cloud? What is the level of the ground?\n\nIt'll take less time for a raindrop to hit the top of a mountain than to hit the ocean.
589948
This is a cool question.  If you were able to tunnel through the Earth, and fall through, your speed would keep increasing until you reached the graviational center.  At that point, the net gravitational force is zero, because all of the forces cancel.  However, you wouldn't just stop because of your inertia.  You would go right through, but then you would begin to slow down, until you reached the other hole.  Depending on how high you started from the first hole, and lets say you jumped five feet above the hole, you would come to a complete stop five feet above the second hole.  Then you would fall back through and continue that cycle.  (Of course we aren't considering air resistance, which screws everything up! ;0)\n\nGuess how long it would take to make it to the other side of the Earth through that tunnel?  Approximately 42 minutes!  Not a bad way of traveling.\n\nThe other cool thing is, if you were to tunnel from say New York to Berlin, which means the tunnel isn't obviously through the middle of the Earth, it would still take you 42 minutes!
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
582615
This is a very good unique question I really don't have an answer but I would guess it may be 50 drops per second?
469350
v = v0 - at\n0 ft = 288 ft/sec - 32ft/sec² t\nt =  288 ft/sec / 32ft/sec²\nt= 9 sec\n\ns = 1/2at²\ns = 1/2 (32ft/sec²) (9 sec)²\ns = 1,296 ft\n\nIt's the same time up and down, so 18sec until it hits the ground.
486427
answer is approxomately 42.476m/s\n\nonly one formula here \nS=ut + (at^2)/2 \ns- distance travelled = 40m\nu- initial valocity\na - acc due to gravity-9.8m/s^2\nt- time \nin the first case the initial velocity is 0m/s^2\nso\n40=(9.8*t^2)/2\nt=2.857 sec\nas both the balls have to touch the ground at same time  and the second ball being dropped( rather thrown would sound correct) after 2 secs, the later has only 0.857 sec left for traelling a distance of 40m. \nso \n40=u*0.857+(9.8*(0.857^2))/2\nsolving u=42.476m/s\nthe initial velocity for the second ball should be 42.476m/s and should be thrown vertically downwards.\nthat is pheta=0 in the formula given un the question.
43880
I'm just doing all your work aren't I?  Its ok I like physics, and its good practice.  I am studying for the MCAT coming up this April (Med School).  This problem unlike the other ones that I did is a little more tricky to me.. Mostly because I don't have my physics textbooks with me.  But it would appear that you need to calculate the acceleration that is exhibited on the foot when it hits the ground.  The way I approached this problem is by using vf^2 = vi^2 + 2ad  and I solved for acceleration.  Knowing that the final velocity is zero I isolated acceleration to get and acceleration of -1024 m/s^2!!! thats huge isn't it.  that is why I'm uncertain about this problem... anyways, multiplying that using the F =ma equation I got a total force of 9216N.  But after inspecting the inital velocity its got to be right because if you convert 6.4 m/s into km/hr you get 23.04 km/hr!!! This person must be falling of a building or something!  So there it is.  Afterall didn't the question start out saying that injury can happen!?