d=-5(t=1)^2 + 258....so d represent  metres in height, t reperesent t...so how long does the object take to reach the ground to the nearest tenth
124442
ok. so you have to find t when d=0. cuz the ground is at d=0.\n\nso d=-5(t-1)^2 + 258\n0 = -5 (t-1)^2 + 258\n0 = -5 (t^2 - 2t + 1) + 258\n0 = -5t^2 + 10t - 5 + 258\n0 = -5t^2 + 10t + 253\nso now you should use the quadratic equation to find what t equals.\n\nt = -10 +- sqrt (100 - 4(-5)(253)) / -10\nt = -10 +- sqrt (5160) / -10\nt = -10 +- 71.1336 / -10\nt = -81.1336/-10     or    t =  61.1336/-10\nt = 8.11336s     or t = -6.11336s.  and time cant be negative so the time it would hit the ground is 8.1s. \n\nhope that helps. laterz
141471
An egg is pretty aerodynamic, so we can neglect friction. Thus, a falling object satisfies the equation\n\ns(t) = -0.5*g*t^2+v0*t+s0\n\nwhere g is 9.82 m/s^2, v0 is the start velocity (in this case, 0) and s0 is the start height (in this case, 3 metres).\n\nSo, letting s(t)=0 and solving for t, we get\n\n0 = -0.5*g*t^2 + 3\n\nt = sqrt(6/g) = 0.78 seconds.
473909
In the vertical direction earth is pulling the object downward with a force.\n\nAnd all objects are pulled down by the earth with an acceleration given by g = 9.8 m/ s^2.\n\nFirst we consider the vertical motion alone.\n\nIn the vertical direction \nthe initial velocity u = 0.\nthe acceleration  a = g. \nThe time t = t to be found.\nDisplacement in the vertical direction   s  =   180 m.\n\nWe use the equation s = ut + 05 a tt.\n\nUsing this we find t = 6 second.\n\nNow let us consider the horizontal motion.\n\nThe object is released from the plane which was moving with a speed of 20m/s.\n\nThere fore all objects which was in the plane will also have a speed of 20m/s in the horizontal direction.\n\nHence in the horizontal direction the initial speed is 20m/s.\n\nThe acceleration in this direction is zero since no force is acting in this direction.\n\nThe time t is the same as 6 second.\n\nDistance we have to find.\n\nWe use the equation s = ut + 05 a tt.\n\nSince a = 0 the equation reduces to s = ut. = 20 x 6 = 120m\n\nThe horizontal displacement is 120 m.
549856
Sand falls onto a conical pile at rate of 10 ft^3/min. The radius of the base of the pile is always equal to 1/2 of its altitude. How fast is the altitude of the pile increasing when it is 5 ft deep?\n\nLet radius = r and altitude = h\nThen r = ½h\n\nV = ⅓πr²h\n= ⅓π(½h)²h\n= πh³/12\n\nNow dV/dt = dV/dh * dh/dt\nSo 10 = πh²/4 * dh/dt\nWhen h = 5\ndh/dt = 10/(π*5²/4)\n= 8/5π ft/min\n≈ 0.51 ft/min\n(≈ 6.1 inches/min)
218060
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
403122
I solved the problem by stating that P would travel distance d in time t\nd=20*t+.5*.3*t*t\n\nand Q would travel d+20 in time t\nt is the time to collision\n\nd+20=30*t\nd=30*t-20\nSubtract the equations and solve for t:\n\n0=.15*t*t-10t+20\nt=2.064 seconds\n(the second root is superfluous)\nd=41.9m\n\nWhich proves collision\n\nj
198058
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)
645545
The height of the trajectory is 1 m (3.05-2.05)\n\nThe horizontal distance is 6.02 m\n\nNow you have ucos(25)t = 6.02\n\nand usin(25)t-0.5*9.8*t^2 = 1\n\nSubstitute t = 6.02/ucos(25) in the 2nd eqn and solv.
403039
dC/dt = 4 m/s\nC = 2*PI*R\ndC/dt = 2*PI*dR/dt = 4\nSo dR/dt = 2/PI\n\ndA/dt = d(PI*R^2)/dt = 2*PI*R*dR/dt = C*dR/dt = 32 * 2/PI = 64/PI m/s
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
533530
t=2.512 seconds\n\n10.3=0.5(0+8.2)t\nTherefore \n\n1 = 4.1t/10.3\n\n1/t = 4.1/10.3\n\nt = 10.3/4.1 = 2.512 seconds
146272
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!!!
247342
Since the height is the only thing considered you can disregard the initial 32ft/sec (this is the initial x direction velocity).  When the height is 12ft you set the equation equal to 12 and use the quadratic formula to solve for t {[at^2+bt+c=0; a=-16, b=32, c=-12] stick these values into    (-b+/-sqrt(b^2-4ac))/2a} You'll find that t=1/2,3/2.  He reaches 12 feet going up at .5 secs and again coming down at 1.5 secs.  Do the same for 14ft and you should get two answers (0.646s,1.35s) He is above 14ft between these times so you subtract the larger time from the smaller and you get about 0.707 secs.\nIf you have a problem understanding my notation check out the quadratic Equation at http://mathworld.wolfram.com/QuadraticEquation.html
261468
T = 2 * pi * sqrt(l/g)\n2 = 2 * pi * sqrt(0.9942/g)\n--- taking your length to be in metres\n\ng = 9.8123607 m/s2
378002
first get the time from the horizontal motion  t = .51 / 1.2 s (uniform motion, no acceleration here)\n\nthen use h = 1/2gt^2 for the vertical motion (with initial velocity zero)
28328
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.
196316
Without regard to air resistance you solve the problem as such.\n\nv(f) = final velocity\nv(i) = initial velocity\na = acceleration\nt = time\nx = displacement (distance)\n\nWhen you throw the ball up vertically, you know several things:\n1) v(i) = 30 m/s\n2) v(f) = 0 m/s\n3) a = -9.8 m/s^2 (this is because acceleration due to gravity goes in the opposite direction of the object)\n\nSolve for x with this equation:\nv(f)^2 = v(i)^2 + 2ax\n0^2 = (30 m/s)^2 + 2(-9.8 m/s^2)x\n0 = 900 m^2/s^2 - (19.6 m/s^2)x\n(19.6 m/s^2)x = 900 m^2/s^2\nx = (900 m^2/s^2)/(19.6 m/s^2)\nx = 45.9 m (rounded off)\n\nNow you know how high the ball went.  The time it takes to reach this is equal to the time it takes to get back down (assuming that it is the same distance up and down).\n\nTo solve for t, use this equation:\n\nv(f) = v(i) + at\n\n0 = 30 + (-9.8)t\n9.8t = 30\nt = 30/9.8\nt = 3.06 s  this is going up to max height.\n\nto check if it will be the same time going down, use this equation:\n\nx = v(i)t + (1/2)at^2\nv(i) = 0 (because when you reached max height, your velocity will be zero)\na = 9.8 because your ball is travelling in the direction of the acceleration.\n45.9 = 0 + (1/2)(9.8)t^2\nt^2 = 45.9/4.9 = 9.367\nt = 3.06 s\n\ndone