How long will this take to hit the ground?
470748
When it hits the ground h = 0 \nHence the equation reduces to\n16t^2 - 70t -3 = 0\nt = (70 plus or minus sqrt(70^2 - 4*16*(-3)))/2*16\n  = (70 plus or minus sqrt(5092))/32\n  = (70 plus or minus 71.35)/32\nt= 4.42 s approx  (note that 70 - 71.35 is not applicable since it is negative)\nAnswer 4.42s
414789
make sure the wheels hit the ground first.
370877
2-3 days I would think:)
72074
h=-(1/2)*g*t^2+Vo*t+ho\n\nh=hieght (meters)\ng=gravity (9.8 meters per second^2 on earth)\nt=time (seconds)\nVo=intial velocity (meters per second)\nho=intial hieght (meters)\n\nyou are right about t=Vo/g, time it takes to hit max height.\n\nif intial height is zero then time it takes to hit the ground is just double the time it takes to reach max hieght\n\nthese are simplfied equations that ignore air restance, relativity, and changing gravity with distance from earth.
759333
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
917751
about 90 seconds, but it depends on elevation (because of atmospheric pressure)
750320
If it takes 6 secs to reach the max, it will take an additional 6 secs to reach your starting point.  In this case it is 2560.  Then it is a matter of finding the time from that point to going to the ground.\n1st lets correct the orginal equations (acceleration due to gravity is down/negative)\nh = -16t^2 + 192t + 2560\n2nd - set the equation to zero since we want to find the time to ground.\n0 = -16t^2 + 192t + 2560\n16t^2 - 192t = 2560\nt^2 - 12t = 160\n(t^2 - 12t + 36) - 36 = 160\n(t - 6)^2 = 196\nt - 6 = 14\nt = 20\nConclusion: 6 seconds to reach max ht of 3136, another 6 seconds to reach starting point of 2560 and another 8 seconds to hit the ground.
978262
Hi!\nIn a very simplified model it would take about 2.833 seconds. In reality it takes a little longer than that because the air exerts some drag, however he falls. The time depends still on it's weight because the drag depends on the air and shape, while G=m*g depends on it's weight, so the mass remains as a variable. Very light objects such as balloons fall very slow (as you probably know).\n100km/h=27.7 m/sec.    time*acceleration= (desired speed) When there is no initial speed. Thus time= v/accel=2.83sec
335818
well...throw?..if you dropped them at the same time they will hit the earth at the same time
882616
Right into the ground
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.
818887
Will be a while before that body falls by gravity
441442
you have your units wrong, do you mean 5m/s or 5m/s^2 and you mean 20m from the ground?\nIf acc=5 when height=20 speed= 14.14 in upwards direction (using 2ax=v^2-u^2)\nIf initial speed is therefore -14.14m/s then at t=1 speed=-4.34m/s so it is still going upwards\nt=2 speed= 5.45m/s it is now going down. Both last results worked out using v=u + at\n(Sorry I assumed you meant 5m/s^2 straight up, you need the angle at which it accelerates aswell)
964449
Depends on what mass, what size due to air resistance\nA ball bearing coming from orbit could be in excess of 5mps, a man falling (not from orbit! but maybe 50,000') about 180mph.
873777
pack it in foam, put a parachute on it... and drop it from the BOTTOM of the 3 story building!!!
1033214
throw it to the floor that should help
128576
depends how high off the ground it is.