another physics question. this one is on parachuting.?
93876
A quick guess without having reviewed your math, try adding g = 9.8 m/sec^2 to your answer and see if that is the answer that is needed.
111044
I'm just doing all your work aren't I?  Its ok I like physics, and its good practice.  I am studying for the MCAT coming up this April (Med School).  This problem unlike the other ones that I did is a little more tricky to me.. Mostly because I don't have my physics textbooks with me.  But it would appear that you need to calculate the acceleration that is exhibited on the foot when it hits the ground.  The way I approached this problem is by using vf^2 = vi^2 + 2ad  and I solved for acceleration.  Knowing that the final velocity is zero I isolated acceleration to get and acceleration of -1024 m/s^2!!! thats huge isn't it.  that is why I'm uncertain about this problem... anyways, multiplying that using the F =ma equation I got a total force of 9216N.  But after inspecting the inital velocity its got to be right because if you convert 6.4 m/s into km/hr you get 23.04 km/hr!!! This person must be falling of a building or something!  So there it is.  Afterall didn't the question start out saying that injury can happen!?
976821
Well, these are just opposite vectors, so you subtract the 50 m/s of the wind from the 100 m/s of the plane.\n\nActually it's : 100 + (-50) = 50 m/s at 90°\n\nE-mail the other other questions to me so I can send you a better response with graphics.\n\nI'll be happy to help :)
299084
If you believe everything you read, a woman parachuted from 50 ft, her shoot did not open and she suffered only broken bones.  It was in our local paper last night.  I personally do not know nor care how high I can jump, I'm just greatful I can walk and not fall down most days!  But good luck in your experimentation = sometimes that's the best teacher.
1087896
At what altitude is the lead bullet being fired??   And what is the rest of your question??
716994
Final Examination Physics 104 19 May 1977 Attention: in this ...\nThe pipe turns vertically upwards as shown in the sketch; the vertical part has a smaller ... Dick, together with his suit, has a mass of 175 kg, and Jane, ...\nwww.iit.edu/~johnsonp/exams104/final190577.html - 31k - Cached - Similar pages\nNewton's Laws Applications Review - Q#13-Q#21 Answers\nSince there are two forces pulling upwards and since the sign is hanging ... (0.250 kg)•ax = Ftens - 0.25 N. Analyzing the Fy forces on the 50-g mass yields ...\ngbs.glenbrook.k12.il.us/academics/gbssci/phys/chemphys/reviews/u8breview/u8ans2.html - 28k - Cached - Similar pages\nNewton's Laws Review\nIf a football is moving upwards and rightwards towards the peak of its ... A 0.250-kg rightward moving air track glider decreases its speed from 0.872 m/s ...\ngbs.glenbrook.k12.il.us/Academics/gbssci/phys/chemphys/reviews/u8areview/u8aprint.html - 32k - Cached - Similar pages\n[ More results from gbs.glenbrook.k12.il.us ]\n[PDF]\nPHYS 1002 Physics 1 (Fundamentals) – Semester 1, 2006\nFile Format: PDF/Adobe Acrobat - View as HTML\nof mass, torque and static equilibrium. You will also learn the concept of momentum, energy, ... 0.250 kg, m. 2. = 0.375 kg, g = 9.81 ms ...\nwww.physics.usyd.edu.au/ugrad/jphys/1002_mod2_mech.pdf - Similar pages\nThe Physics Classroom\nNote that the vertical forces balance but the horizontal, forces do not. The net force is 129.9 N, right - 60 N, left. The mass is 20 kg (Fgrav/g); ...\nwww.physicsclassroom.com/Class/vectors/U3L3d.html - 31k - Cached - Similar pages\n[PDF]\n∑ ∑\nFile Format: PDF/Adobe Acrobat - View as HTML\nvalue in the upwards direction. (b) For no vertical acceleration we have: ... 6 A ball having a mass of 0.250 Kg is held at point A against a spring which ...\nwww.physics.villanova.edu/faculty/shaw/2400/tests/s97all4.pdf - Similar pages\n[PDF]\nPhysics 2400 Test #2 (8:30 Class) Spring 2002\nFile Format: PDF/Adobe Acrobat - View as HTML\n3 (a) A boy's toy cannon fires a 0.250 (kg) rubber ball vertically. ... When the cannon is fired, compressed gas below the ball exerts an upwards force, F ...\nwww.physics.villanova.edu/faculty/shaw/2400/tests/s02all2.pdf
901641
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
484307
in air= 340m/s\nin water=1500m/s\nin solid=4500m/s
951127
first why do we need his moving 30 meters and what does it mean anyway? i don't understand the question here.. first he let the kite and it was flying and the string made an angle of 47 degrees.. and then he walked somewhere 30 m? so now it is an angle of 39 degrees? i could say some little things to help only:\n\n-we watch a man as a spot that doesn't move\n-a string has first brace at force T1, and it is equal the strength of wind that is flowing against kites dimension S.. so it is T1=S(paralel) * P(paralel).. \n-this task is too complicated, i'll write it down, see your mail tomorrow and send you, it is 2 am here..\n-good night
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.
728806
flying faster than the gravitational pull that is pulling you back
921378
Projectiles without the presence of air pressure creates a parabolic trajectory (path taken by a projectile).\n\n^_^
1124515
Since the plane flies by producing thrust and thereby offsetting its weight, the weight in flying state is zero. The mass remains the same as it was on the ground.
446591
1)X=V1t+1/2gt^2\nat the interval between getting to the highest point and being dropped back to the first place is half of the full time = 2 seconds\nV1 at the highest point = 0\nX=displacement\nX=0+1/2*9.8*4=19.6 m \n2)V2=V1+gt\nV2 = velocity of landing\nV1 = velocity at highest point\nV2 = 0 + 9.8 * 2 = 19.6 m/sec
767928
1. its final velocity depends on the direction it was heading to start off with. If it was flying in the x direction (no components in y or z) then after 4.3 seconds the x component has increased by 8.92*4.3  and the y component has increased (from 0 this time) by 1.21*4.3.\nso now you can work out the new direction (in the x direction but slightly downwards by whatever angle has a tangent of 1.21*4.3/(152+8.922*4.3) \nand using pythagoras to determine the speed.\n\n2. You can ignore the gravity as it is << the precision with which the acceleration due to the electric field is stated (I'm assuming the 106 is 10^6 = 1000000) \nso the final velocity is 3.80*2.45*1000000 m/s upwards.
170531
1) V2^2=V1^2+2d*g ----> 0=47^2+2d*-9.8 ----> d=112.7 m\n2)remeber when mesuring the time we take the velocity on the Y axes and when mesuring range or distance we take velocity on the X axes got it?\nd=V1+0.5*g*t^2 and since the rock is thrawn horizontaly the velocity on the Y axes once it's thrawn is zero, by the way the decplacement would be minus cuz the rock is projecting downwards \n-3.5 m = 0 + 0.5 * -9.8 * t^2 -----> t=0.845 seconds\nd=V*t=47*0.845= 40 m\n3)this is a tough one you'd better draw it :\na.V2^2=V1^2+2d*g ----> 0= (47sin37)^2 + 2d*-9.8 ----> d=40.8m\nb.V2=V1+g*t ---> 0=47sin37+-9.8*t ---> t=2.886 seconds\nrange=V*cos37*t=47cos37=108 m\nc.V2=V1+g*t=47sin37+(-9.8*2)=8... m/sec (velocity on the Y axes)\nvelocity on the X axes is constant and equals=47cos37=37.5m/sec\nthe net velocity^2=37.5^2+8.685^2 ---->velocity of rock after 2 seconds of launch=38.5m/sec
1069779
Draw a free-body diagram of the ball and examine it's accelerations - if you assume wind and air resistance are negligible and that the ball is accelerated straight up, then you're left with its initial acceleration (68.57 m/s^2) and gravity (9.8 m/s^2 if I remember correctly) \n\nThe ball will decelerate by 9.8 m/s every second, you can divide to find the amount of time the ball takes to reach its apogee (point of zero acceleration), then integrate the acceleration curve to find the velocity curve, which can be integrated again to find the distance traveled.  I'm not sure it can be done without the calculus.