How fast will this be going once it hits the ground from a 30 foot drop?
443173
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.
861868
what's the 30 g for? here's my answer:\n\nfrom F=m*a,\n\nF= (70 kg)*(3m/s^2)\nF=210 newtons...\n\nthis is the force needed to help the 70 kg person survive the crash...does that help?
435954
i'm assuming that you are ralking bout a hole being drilled right thru the earth and somebody falling into it...............you would accelerate and move towards the center and reach the opposite end of earth, at which point your velocity will reduce to zero and you will again move back to the cetre of earth and then onto the starting place.............in simple words you would execute Simple Harmonic Motion(SHM)
747360
Well gosh....  \n\nsince the only linear unit mentioned in the question is 18 ft/s  let's assume that the 60 is going to give us acceleration in ft/sec^2 \n\nSo, the position function for this is the second integral of the acceleration function, right? \n\nso...  the first anti-derivative of 60t is  30t^2   which gives the velocity, so at the end of the boost phase, our rocket is moving 30 * 3 * 3 = 270 ft/s \n\nand the second antiderivative is 10t^3  right?  \n\nSo, the position at time T is 10 * 3 * 3 * 3 = 270 feet.  (odd coincidence, but there you are...) \n\nNow, the rocket begins a SECOND acceleration by gravity of -32 ft/s negative 'cause we're accelerating DOWN now... ....   but this time the acceleration is constant,  so \n\nV = gt   and s = 1/2  a  t^2 so, we're slowing from 270 feet per second to zero at the top...  and  t = v/g = 270 / 32 = 8.44 seconds and the height we coast to is the 270 we had at the end of boost plus   1/2  *  32  * 8.44 * 8.44 = 1140 feet.  \n\nThe parachute is going to open after 14 seconds...   so, 14-8.44 is 5.56 seconds of free-fall (and we'll ignore air resistance again) so, we go from zero to something in 5.56 seconds....   and position is 1140 - ( 1/2  * g  * t^2 ) again...   so, the parachute opens at \n\n1140 - (1/2  * 32  * 5.56 * 5.56 )  = 645 feet. \n\nad the velocity is  gt = -32 * 5.56 = -178 feet per second...  \n\nNow, for the NEXT five seconds, we slow under the effect of the parachute   to  -18 ft/sec -- so,  we slow by 160 feet per second in five seconds, so the acceleration of the parachute is (oddly) 32 feet/second squared (huh, how odd is that? )  \n\nSo, the distance traveled during that phase is Vo t + 1/2 at^2  \n\nSo, we have (-160 * 5) + (1/2*32*5*5) =  -800 + 400 = 400 feet.  \n\nand from 400 feet up, we continue down at 18 ft/s for an additional  22 seconds...  \n\nand the rocket lands at t = 22 + 5 +  14 + 3  = 44 seconds \n\nAnd yes, I dropped a bunch of decimal points.  Your original data was only to two significant digits....
827544
1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj
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.
167921
increase the drop distance.\nthe further you let it fall before stopping the longer it will take, or you could try an updraft
1140237
125 mph  it's the same for all things dropped
544074
About 30m (100 ft) high.
534824
60mph is 88 ft/sec; (memorize that).\nSo, 90mph is 132 ft/sec.\n\nIgnoring air friction (which DOES have an effect!), it takes 90/132 of a second for the ball to travel that far - or about .682 seconds.\n\nHow far does something drop in about 2/33 of a second? Several feet.
128576
depends how high off the ground it is.
651651
tip it over, just make sure you use medium force back up about 10 feet and run as fast as you can to the center of its body.
102851
yes it would, even considering the acceleration of gravity. you still have to factor in terminal velocity. The mass of the ant compared to its area and wind resistance would limit it's speed during it's fall. After approximately 10ft, it wouldn't gain anymore speed. Quite possibly, with the wind, it would land several blocks away.
520357
Try it at different speeds,\nEx:-\nFirst try at 2 m/s\nThen try at 5 m/s\nThen try at 10 m/s\nThen if you survive, try at 20 m/s\nTry using concrete wall if you want to have more fun :)
867531
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!
788979
For the add on: If a toy plane weighs two pounds, how fast should it accelerate? What unit of measurement would you use?\n\nNot sure what you're asking. Unless you're talking about how fast it would accelerate if you drop it. Then it would be vertically down at 32 ft/s^2 in a vacuum. No matter how much it weighs. (I'm using the English system units since you talked about pounds.)\n\nIf you're asking for an opinion of what would be a nice number for acceleration ... Well, if you wanted it to get up to a speed of 50 ft/s within 20 seconds:\nV = a*t or acceleration = V/t = (50 ft/s)/20 s = 2.5 ft/s^2\n\nI hope this helps
1143368
Change the question to "fall in air" and think about it. A parachute drops slower than a rock. The shape influence the speed due to drag.\n\nSame for water.