For 1989 and 1990 DaVE johnson had the highest decathlon score in the world.When Johnson reached a speed of 32 ft/sec on the pole cault runway, hs hieght above ground t seconds after leaving the ground was given by h=-162^2+32t. For how long was he more than 14 ft in the air? \nPlease show your work
203007
I think your equation should say “h = -16t^2 + 32t” - the “-162^2” part doesn’t make sense.  He goes up and down following a parabola, so there has to be a t^2 term somewhere for this equation to work.\n\nI’m going to assume that it should be -16t^2 and go from there.  To find the point where he gets to 14 feet, simply set the equation to 14:\n\nh = -16t^2 + 32t\n14 = -16t^2 + 32t\n0 = -16t^2 + 32t – 14\n0 = 8t^2 - 16t + 7\nt = [16 +/- sqrt((-16)^2 - 4(8)(7))] / 2(8)\nt = [16 +/- sqrt(256 - 224)] / 16\nt = [16 +/- sqrt(32)] / 16\nt = 1 +/- sqrt(2)/4\nt = 0.65s, 1.35s\n\nSo at 0.65s and 1.35s, he was at 14 feet.  Subtract the two and you find that he was at or above 14 feet for 0.7s.  (For a precise time, you can subtract [1 + sqrt(2)/4] - [1 - sqrt(2)/4] and get sqrt(2)/2.)
495810
Margaret is in 1st. Let's say she's at the 75 meter mark.  Manuel is 30 behind her, so he's at 45. Now, assuming Coreen is at the 0 mark (75 behind Margaret), then Tom is at the 20, or 20 ahead of her. You take Manuel's mark (45) and subtract it by Tom's (20) to get 25.\n\nHope that helps!
443210
This question can be answered with both calculus and standard constant acceleration equations.  In constant accelaration, velocity equals acceleration multiplied by time (v = at). Taking the integral of this gives us popular phyics equation d = v0t + (1/2)at^2, or distance equals initial velocity multiplied by time plus half the acceleration multiplied by the time squared.  Using these equations we have:\n\nAt peak of the dive:\nv = v0 - at\n0 = 12 - 9.8t\n9.8t = 12\nt = 1.22 (time in the air at peak)\n\nHeight at peak:\nd = vot + (1/2)at^2\nd = 12(1.22) + (1/2)(-9.8)(1.22)^2\nd = 7.34684 feet\n7.34684+10 feet (height of board) = 17.34684 feet above water\nSo, the answer to A is 17.347 feet\n\nWhen she will hit water:\nd = (1/2)at^2\n17.34684 = (1/2)(9.8)(t^2)\nt^2 = 3.54017\nt = 1.8815 (falling) + 1.22 (to reach peak) = 3.1015 seconds\nSo B is 3.1 seconds\n\nShe was falling for 1.8815 seconds at 9.8 m/s, so using v = at we have:\nv = 9.8(1.8815)\nv = 18.4387 m/s\nSo C is 18.4387 meters per second.
332471
About 1000 feet per sec.
380436
Probably from 6 to 10 feet., if the drop is feet first, and you don't get the air or wind knocked out of you when you hit. I was about that size on my first hi dive jump. Although they usually measure it in meters. Chances are, when you look at it from that height, you may decide otherwise unless provoked by someone in a dare! the distance after about 30' doesn't change as the speed rate of fall will be the same.
46934
First of all, welcome to the UK! As you may recall, London won the bid to host the 2012 Olympics (and as I would like to get involved in some form or another perhaps I might just bump into you, not literally mind). Anyway, on to your questions:\n\nMmm, well let's have a look at who won gold in Athens in 2004 for the US and see if we can get some pointers. \nJustin Gaitlin (100m), 6ft 1 and 180lbs, born 1982\nShawn Crawford (200m), 5ft 11 and 165lbs, born 1978\nJeremy Wariner (400m), 6ft and 155lbs, born 1984\nTeam USA (4x400m Relay), HARRIS, Otis, 6ft 1 and 165lbs, born 1982\nBREW, Derrick, 6ft 1 and 175lbs, born 1977\nWARINER, Jeremy, 6ft and 155lbs, born 1984\nWILLIAMSON, Darold, 6ft 2 and 170lbs, born 1983\n\nSo on average:\n\nHeight: 72.57 inches (6ft 0.6 inches)\nWeight: 166.43 lbs\nBirth Year: June 6th 1981 at 10.17am local time!
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)
471316
deceleration = change in speed / time\ndeceleration = (50 (km / hr)) / (0.15 s) = 92.5925926 m / s2\n\nF = ma\nF = (1 500 kg) * 92.5925926 (m / (s^2)) = 138 888.889 newtons
140364
By double click do you mean the hitch kick?? There was an american sprinter/long jumper called Jesse Owens from ohio who, in 1935 set a world record of 8.13m, it stood for 25 years before being broken again. I think the hitch kick just evovled, it is adopted and modified by every male jumper and selected females (however they tend to employ the hang technique).
536926
s = 1/2 at^2\n40 m = 1/2 * 4 m/s^2 * t^2\n80 m / 4 m/s^2 = t^2\n20 = t^2\nt = square root(20) = 4.47213595 seconds to go the 40 meters \n\ndinosaur going (4.47213595 seconds) * 4 (m / (s^2)) = 17.8885438 m / s at that time\n\nabout another 2 seconds to get to max speed of 25 m/s
934
You're really close! The answer is actually 1:03:11. You were off one minute because you forgot to borrow a minute from the 34 so you could subtract the seconds.
207642
Popov has the record, around 21.6 seconds I believe\n\nfor future reference.. its:\n\nAlex Popov\nMichael Phelps\nIan Thorpe
306062
Michael Johnson - 200 m - 19.32s (9.66s per 100m)
521270
ok my friend\ni assume that by now you know that maximum distance is achieved by throwing something at 45 degrees. so start by that and take into consideration that Ux is constant and is Ucos45\nso \nt = 75 / Ucos45\n\n\nnow at 75meters the Height is zero for Uy component\nmeaning that \n\n0 =H = Uyt - 1/2gt^2 =>\n\nusin\n\nUsin45 *  = 1/2 g * 75/Ucos45 ==>\n\nU = sqrt ( 75*9.81) => U = 27.12 m / sec\n\ni used the fact that time is the same and for Ux the distance is 75 but the same time leads to height zero for Uy.\n\n\nnou UX = S/t=>\nt = 75 /Ucso45* \n\nand t=3.91sec
272804
http://www.answers.com/What%20is%20the%20toughest%20competitive%20examination%20in%20the%20world%3F
443265
1. v = d/t\nt = d/v\nt = (3 900 km) / (299 792 (kilometers per second)) = 13.0090196 milliseconds\n\n(150 000 000 km) / (299 792 (kilometers per second)) = 8.33911512 minutes\n\n (444 000 000 000 km) / (299 792 (kilometers per second)) = 17.1415144 days = 411.396346 hours\n\n\n(10.34 hours) * 299 792 (kilometers per second) = 1.11594574 × 10^13 meters
671139
The distance between the start and finish can be considered the hypotenuse of a right triangle with the 35 and 65 km as the two legs ( I assume you're in the right triangles chapter)\n\nPythagorean theorem says :\n\n(Leg1)^2  + (Leg2)^2 = (Hypotenuse)^2\n35^2  + 65^2 = (hypotenuse)^2\n1225  + 4225 = (hypotenuse)^2\n5450 = (hypotenuse)^2\n73.8 km = hypotenuse = distance from starting point