You accidental throw your car keys horizontally at 8.0 m/s from a cliff 64-m. high. How far from the base...?
956205
First compute the fall time of the keys.  Since you threw them horizontally, the fall is due only to gravity:\n\nd=1/2*g*t^2\n64=.5*9.8*t^2\nt=sqrt(2*64/9.8)\nonce you have t, just multiply the horizontal velocity times the time of flight to get the distance traveled:\n\nd=v*t\nd=8*sqrt(2*64/9.8)\n=29m\n\nIf you didn't understand how to compute the fall time, here's a bit more info:\n\nThe average velocity is\n1/2*g*t\ncompute the fall distance as\n1/2*g*t*t\n\nj
961997
1.7 kilometres. Move the decimal three places to the left.
1176247
It would never reach the street.  Even if you threw it horizontally as hard as you could it would not hit the street.  The lower floors of the building have setbacks, so the upper floors are set in quite a distance from the street.  It would land on the roof at least several stories up.
1149306
I think you can use a simpler equation than the one for distance.\nWe don't need to find distance, just time.\n\nv = u + a.t\nv= 0, u= 64 f/s, a= -32f/s^2\nsolve for t. \nt = 2 s\nThats the time it takes to go up just before it starts to fall again.\nThe time taken to fall is going to be the same: another 2 s\n\ntotal time: 4 s
240171
i dunno im thinking about 100 ft  high i dunno ya met wanna ask someone that is smarter than me it would probably range about 100 ft - about 10000000000ft  i dunno
1008231
Set the equation of motion.\nTaking upward direction as positive\ns=vt-1/2gt^2\nat maximum height s=10,v=0\ng=-g since upward direction is positive\nyou get t= sqrt(2s/g)\n=sqrt(2*10/9.8)\n=1.428 secs
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
759984
The total time for the rock to drop and the sound to come back is 2.40s.  The distance that the rock drops is (.5)(9.8 m/s^2)(time for rock to fall).  The distance that the sound comes back is (336 m/s)(time of sound to come back).  The two times added together are 2.4.\nThese two distances are the same so.\n(.5)(9.8m/s^2)(time for rock to fall)^2= (336 m/s)(2.4 s - time for rock to fall) \n(4.9m/s^2)(time)^2 = 806.4m - (336 m/s)(time for rock to fall)\n(4.9m/s^2)(time)^2 +336 m/s(time) -806.4 m = 0\nUsing the quadratic formula, the allowed values for time is\n2.31 s\nDistance down to water is (.5)(9.8 m/s^2)(2.31s)^2 = 26.15 m\n\nb) if you were to ignore travel time for the sound, you would simply have (.5)(9.8m/s^2)(2.4s)^2 = 28.22 m.  Your percentage error would be (28.22-26.15)/26.15 = 7.9%
547462
I have a suggestion:\n1) go outside at exactly midnight and place your thumb on the exact center of you left rear wheel/hubcap for exactly 32.5 seconds\n\n2) stand due North and blow into the wind at for exactly 15.7 seconds\n\nAfter doing this your car doors should open up and you will be able to retain your keys.  Good luck!!!
970380
Let rock be thrown at an angle of θ at a velocity of Vo\n\nThen vertical component = Vo sinθ\nand horizontal component = Vo cosθ\n\nAlso acceleration in horizontal direction = 0 \nand acceleration in vertical direction = -g = -9.8 m/s²\n\nVertical direction\n\na = -9.8\nv = u + at\nSo Vt = Vosinθ - 9.8t\nand s = ut + ½at²\nSo h = Vosinθ*t - 4.9t²\n= 0 when it is at ground level\nie t(Vosinθ - 4.9t) = 0\nSo t = 0 or Vosinθ/4.9 ... ****\nSo it will take Vosinθ/4.9 s to cross the river.\n\nHorizontally\n\nVt = Vocosθ + 0 (as a = 0)\n= Vocosθ\n\ns = ut + ½at²\n= Vocosθ * t\n\nSo 75 = Vocosθ *Vosinθ/4.9\n= Vo².sinθcosθ / 4.9\n= Vo² * 2sinθcosθ / 2*4.9\n= Vo² * sin2θ/9.8 \nThus Vo² * sin2θ = 75*9.8 = 735\nThus Vo² = 735/sin2θ\nNow least value of Vo occurs when Vo² is minimum and this will occur when sin2θ is maximum (=1)\n\nSo Min possible Vo = √735 \n≈ 27.11 m/s (projected at 2θ = 90° ie θ = 45°)\n\nMinimum time = Vo*sin45°/4.9 (from ****)\n≈ 3.91 s
479843
I'm assuming it's okay to neglect air resistance in this problem, and that the vertocal velocity component you requested is right before the thingy (I love the word "thingy") hits the ground.\n\nWe have the kinematic equation: S=160+V(initial)t+.5at^2.\na is -32 ft/sec^2, and the initial vertical velocity is 0, so we have S=160-16t^2.  When it hits, S=0, so:\n\n160-16t^2=0 --> 16t^2=160  -->  t^2=10  -->t=10^(1/2)\n\n(Or t = (square root of ten) seconds.  About 3.16 seconds.)\n\nNow that we have t, we just multiply that by the horizontal velocity:\n\n960*10^(1/2), or 960 square root of 10.  (In feet.)  \n\nTherefore the object will land 960*10^(1/2) feet from the base of the building.  Roughly this is about 3035.78 feet.\n\nTo get the final vertical velocity, we use the equation:\n\nV(final)^2=V(initial)^2+2a delta S.  \n\nChange in S in the vertical direction is -160 ft, and a=-32 ft/second squared, and V(initial)=0, so:\n\nV^2=2(-32)(-160)=10240 ft squared per second squared.\n\n--> V=32*10^(1/2), or about 101.19 feet per second.\n\nI hope that's correct.  I typed it with one hand while not really paying attention.  ^_^  From now on, do your own homework.  :P
1118371
double_nubbins has the right approach.  I will add a couple of hints: how long will it take the car to fall 19ft (remember the acceleration equation s=.5*g*t^2).  Assuming there is no air drag, the horizontal velocity of the car must be enough to cover the 52m width in that time.\n\nSpeed of the car? The horizontal speed? The falling speed?  The speed in the direction of motion?\n\nHorizontal speed: as calculated above\nFalling speed = g*t, where t is the time calculated above.\nSpeed  in the direction of motion = sqrt(horiz speed ^2 = falling speed ^2)
824167
The radius is actually 2/3 of the altitude.  Try that
831310
It seems the 16 m/s is not relative to this question. That is velocity in the horizontal direction. Gravity only effects in the verticle in this case. \n\ntime = sqrt(2 x height/ g) = sqrt(2 x 45 x 9.8) = 3.03 seconds\n\nvelocity = time x acceleration of gravity = 3.03 x 9.8 = 29.694m/s
27693
The deepest point is 560 feet (170 meters).  By the way, you're in WAY the wrong catagory.
1135961
in order to solve this problem, you need to work part (b) first before (a)\n\nb): you know gravity (g=32.2 ft/sec) acts opposite the rocket, so calculate, using:\n\nv_final = v_initial - 2*g*y_1\n\nyou know that the final velocity is zero (max height) and you are given the inital velocity. the negative sign is to account for the opposite direction of gravity. move the equation around to solve for y_1 (height from tower to max. height). then add 100 feet (to account for the tower) and you have your answer.\n\na): since you know your maximum height (from part b), substitute that answer for h. [*NOTE: remember to subtract 100 feet from your answer in part b, because t=0 at the top of the tower] You now have a quadratic equation and solve for t using the quadratic formula (you should know what that is). your answer will be the answer!\n\nThis prob is easy only because it worries upto the max height. if we had to worry about the falling down part, well thats another story.\n\n** i already worked out this prob, so if you want to ask more questions/verify your answer, just post an additional response and ill modify my question to answer it. GL!
167582
A friend of mine thought it would be fun to do this.  He  put the transmission in neutral, turned the engine off and removed the keys from the ignition while he was going about 45mph.  This was OK until he approached a sharp turn and dropped the keys under the seat while fumbling to get them back in the ignition.  When he began to turn the wheel without the keys the steering column lock engaged and he ran off the road and hit a large snowbank which stopped him and probably saved his life.  DO NOT EVER try removing the keys from a car in motion. It is suicidal.